Sovle 6x^2+10x+7=0?

Answer 1

x = [ - 10 ± √(100 - 168) ] /12
x = [ - 10 ± √(- 68) ] / 12
x = [ - 10 ± i 2√(17) ] / 12
x = - 5/6 ± i (1/6)√(17)
x = (-1/6) (5 ± i √17 )

Answer 2

Note to the guy a few answers below me: that's the quadratic forumula, not the Pythagorean theorem.

I'm feeling saucy today, so I'll use completing the square.

Divide by the x^2 coefficient. Move the constant to the other side.
x^2 + (10/6)x + (7/6) = 0
x^2 + (5/3)x = -(7/6)

Now take half of the "x" coefficient, square it, and add it to both sides.
x^2 + (5/3)x + (5/6)^2 = -(7/6) + (5/6)^2

Complete the square, take the square root, simplify
(x^2 + (5/6))^2 = -(42/36) + (25/36)
You get a square that equals a negative number, so there is NO REAL SOLUTION.

You can also see by the discriminant that b^2 - 4ac < 0.

Answer 3

Hai enziguri2000,

Such equations which cannot be factorized are solved by finding the roots of the equation.

6x^2+10x+7=0, is a quadratic equation in x.

Any quadratic equation will have two possible roots.

The roots of a quadratic equation ax^2+bx+c=0 are given by:

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

In the equation 6x^2+10x+7=0, a=6, b=10 & c=7

Therefore the roots of 6x^2+10x+7=0 are:

x= {(-10)+-[sqrt(10^2-4*6*7)]}/{2*6}
= {(-10)+-[sqrt(100-168)]}/{12}
= {(-10)+-[sqrt(-68)]}/{12}
= {(-10)+-[sqrt(-4*17)]}/{12}
= {(-10)+-(2)(i)[sqrt(7)]}/{12} .....where i = sqrt(-1), an imaginary number.
= {(-10)+(2i)[sqrt(7)]}/{12} and {(-10)-(2i)[sqrt(7)]}/{12}

Answer 4

you need to use the quadratic formula

(-b +/- (b^2-4ac)^0.5)/ 2a

where a=6
b=10
c=7
(-6+ (100 - 4*6*7)^0.5)/36


(100 - 4*6*7)= -68 ---- no real solutions because this part of the equation is negative

but seeing as the discriminate is negative you cannot solve the problem without using complex numbers and im not sure if you have done that sort of maths yet doing

Answer 5

No real answer. Graph it if you're not sure. The parabola never crosses the x-axis at y=0.

Answer 6

there are two values of x since you have to use quadratic equation:

x = (-5 + sqrt(17)i)/6

and

x = (-5 - sqrt(17)i)/6

Answer 7

use the quadratic equation to solve for x

For ax^2 + bx + c = 0

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

Answer 8

they are hopefully looking foor roots

use the pothagarian theorum -B +- square root b^2- 4ac over 2a

-2 +- square root b^2 - 4(6)(10)
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2(6)

Answer 9

You can't make me