how would you solve this equation??

Question

Solve.

(x + 2)(x - 3) = x^2 - 11




and




3x^2+18x - 2 = 0




Please show steps.

Answer 1

( x +2 )( x -3 ) = x² - 11
x² +2x - 3x -6 = x² -11
-x = -5
x =5


3x² + 18x -2 =0
a² + bx + c =0
x = -b±√ (b² - 4ac )/ 2a
x = (-18) ± √( (18)²- 4(3)(-2) ) / 2(3)
x = (-18 ± √348) / 6
x = -3 ± ( (√87) / 3 )

Answer 2

1)

(x + 2) ( x - 3) = x^2 - 11
a) Do the operation of the left side
x^2 - 3x + 2x - 6 = x^2 - 11
b) Combine like terms in the left side

x^2 - x - 6 = x^2 - 11
c) Transpose x^2 to the left side
x^2 - x^2 - x - 6 = x^2 - 11
-x - 6 = - 11
d) Transpose -6 to the right side
-x - 6 + 6 = - 11 + 6
- x = -5
e)Multiply both sides by - 1

x = 5
=======================
2) 3x^2 + 18x - 2 = 0
Since you cannot get a perfect square, Use the Quadratic Formula to solve for x
Quadratic formula =

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

a = 3 ; b= 18 ; c = -2

x = ( - 18 +/- sqrt (18)^2 - 4(3)(-2)) / 2(3)

x = ( - 18 +/- sqrt 324 + 24)/ 6
x = ( -18 +/- sqrt 348)/6
x = (-18 +/- 18.6548) 6
x = (-18 - 18.6548)/6
x = -36.6548/ 6
x = - 6.1091 ===> first root

x = ( -18 + 18.6548)/6
x = .6548/6
x = .1091 ===> second root

Answer 3

( x +2 )( x – 3 ) = x² - 11
x² + 2x – 3x – 6 = x² – 11
– x = – 11 + 6
– x = – 5
x = 5

3x² + 18x – 2 =0
a² + bx + c =0
x = – b±√ (b² – 4ac )/ 2a
Putting the values in above eq.
x = (–18) ± √( (18)² – 4(3)(–2) ) / 2(3)
= (–18 ± √348) / 6
= (–18 ± 2√87) / 6
= – 3 ± [ (√87) / 3 ]
= – 3 ± [ (√87) / 3 ]
= – 3 ± [ (9.33) / 3 ]
– 3 ± 3.11

Answer 4

(x + 2)(x - 3) = x^2 - 11

=> x²+2x-3x-6=x²-11

=> -x-6=-11

=> x=5




3x^2+18x - 2 = 0

ax²+bx+c=0, D=b²+4ac ,x=(-b-squareroot(D))/2a and x=(-b+squareroot(D))/2a

=> D = 18²-4*-2*3 = (20-2)*(20-2)-4*-2*3 = 20²-40-40+4+4*2*3 = 324+24 = 348,

squareroot(D) = squareroot(348)

=> x = (-18-squareroot(348))/6

or x = (-18+squareroot(348))/6

Answer 5

Question 1
x² - x - 6 = x² - 11
x + 6 = 11
x = 5

Question 2
3 x ² + 18 x - 2 = 0
x = [- 18 ± √ (18 ² + 24) ] / 6
x = [- 18 ± √ (18 ² + 24) ] / 6
x = [- 18 ± √ (348) ] / 6
x = [- 18 ± 2 √ (87) ] / 6

Answer 6

1.
(x+2)(x-3)=x^2-11 => x^2-3x+2x-6=x^2-11 =>-3x+2x-6=-11
=> -x=-11+6 => x=5

2) ax^2+bx+c=0 => x=-b+/- sqrt(b^2-4ac)/2a

From the above we have :

x=(-18+/-sqrt(18^2-4*3*(-2)))/6 = (-18 +/- 18.655)/6 =>

x= 0.655/6
x=-36.655/6

Answer 7

x^2-3x+2x-6=x^2-11
-x-6=-11
x=5

delta=b^2 - 4*(a*c)=(18^2)-(4*3*-2)=348
x=[-b+(delta)^(1/2)]/2a x=[-b-(delta)^(1/2)]/2a
x=(-18+(348)^(1/2))/6 x=(-18-(348)^(1/2))/6