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2007-12-22 04:23:56 · 14 answers · asked by delightfulldee71 1 in Science & Mathematics Mathematics

14 answers

x = [- 6 ± √196 ] / 2
x = [- 6 ± 14 ] / 2
x = 4 , x = - 10
OR
(x + 10)(x - 4) = 0
x = - 10 , x = 4

2007-12-25 07:27:17 · answer #1 · answered by Como 7 · 2 0

x² + 6x - 40 = 0

roots of the equation = (-b ± √(b² - 4ac) /2a where a, b and c are coefficients of x² , x and the constant.

a = 1
b = 6
c = -40

(-6 ± √((6)² - 4(1)(-40)) /2(1)
(-6 ± √((36 +160)) /2
(-6 ± √(196)) /2
(-6 ± 14) /2

roots are (-6 + 14)/2 and (-6 -14)/2
= 8/2 and -20/2
= 4 and -10
~~~

2007-12-22 04:30:55 · answer #2 · answered by A Little Sarcasm Helps 5 · 0 1

Hey there!

Here's the answer.

-b±sqrt(b^2-4ac)/2a --> Write the Quadratic Formula.
-6±sqrt(6^2-4(1)(40))/2(1) --> Substitute 1 for a, 6 for b and -40 for c.
-6±sqrt(36+160)/2 --> Simplify.
-6±sqrt(196)/2 --> Add 36 and 160.
(-6±14)/2 --> Evaluate sqrt(196).
-3±7 --> Reduce the fraction.
-3+7 or -3-7 --> Replace ± with a + and a -.
4 or -10 Simplify each expression.

So the answer is x=4 or x=-10.

Hope it helps!

2007-12-22 04:32:36 · answer #3 · answered by ? 6 · 1 2

6^x2+8x+1=0 With comparison to the standard form of ax^2+bx+c=0 we have: a=6 b=8 c=1 delta=b^2-4a= 64 - 24= 40 x=(-b+-sqrt(delta))/2a= (-8-+2sqrt(10))/12 => x1= {sqrt(10)-4}/6 = - 0.13962 x2= {-sqrt(10)-4}/6 = - 1.19371

2016-04-10 12:55:57 · answer #4 · answered by Anonymous · 0 0

x^2 + 6x - 40 = 0
(x + 10)(x - 4)=0
x= -10 and x=4

2007-12-22 19:19:10 · answer #5 · answered by Santiago 3 · 0 1

x^2 + 6x - 40 = 0 can be easily factored by noting that factors of 40 (like 10 and 4), can subtract to give 6.

(x + 10)(x - 4) = 0
so x = -10 and x = 4 are your two roots of the equation.

2007-12-22 04:33:42 · answer #6 · answered by Charles M 6 · 0 2

Given:

x^2 + 6x – 40 = 0

product, ab =-40
sum, a + b = +6
Therefore a= +10 and b = -4

x^2 +10x - 4x– 40 = 0

x(x +10) -4 (x + 10) =0

(x - 4) (x + 10) =0

=> x =4 or x=-10

all the best

2007-12-22 04:30:58 · answer #7 · answered by Roslyn** luv maths 2 · 1 4

a = 1
b = 6
c = -40

Quadratic formula = (-b ± √(b²-4ac))/2a

Now, plug in the values for a, b, and c:

(-6 ± √(36 + 160))/2
(-6 ± √(196))/2
√(196) = 14
(-6 ± 14)/2
-3 ± 7
So the answers are -10 and 4

2007-12-22 04:30:33 · answer #8 · answered by Anonymous · 0 2

x^2 + 6x - 40 = 0
x^2 +10x - 4x - 40 = 0
x(x+10) - 4(x+10) = 0
(x+10)(x-4)=0
x = -10 and x = 4

2007-12-22 04:32:37 · answer #9 · answered by ▐▀▀▼▀▀▌ ►MARS◄ ▐▄▄▲▄▄▌ 6 · 0 3

x2 + 6x – 40 = 0

x = [-b +- sqrt(b^2 - 4ac)]/2a

x = [-6 +- sqrt(36 + 160)]/2

x = [-6 +- sqrt(196)]/2]

x = [-6 +- 14]/2

x = -3 +- 7

x = -10, 4

2007-12-22 04:32:38 · answer #10 · answered by Anonymous · 0 2

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