Question

What are the zeros of the quadratic function f(x) = 6x2 12x – 7?

Answer 1

we have

$f(x)=6x^{2}+12x-7$

Equate the function to zero

$6x^{2}+12x-7=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$6x^{2}+12x=7$

Factor the leading coefficient

$6(x^{2}+2x)=7$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$6(x^{2}+2x+1)=7+6$

$6(x^{2}+2x+1)=13$

Rewrite as perfect squares

$6(x+1)^{2}=13$

$(x+1)^{2}=13/6$

Square root both sides

$x+1=(+/-)\sqrt{\frac{13}{6}}$

$x=-1(+/-)\sqrt{\frac{13}{6}}$

$x1=-1+\sqrt{\frac{13}{6}}$

$x2=-1-\sqrt{\frac{13}{6}}$

therefore

the answer is

The zeros of the quadratic equation are

$x1=-1+\sqrt{\frac{13}{6}}$

$x2=-1-\sqrt{\frac{13}{6}}$

Answer 2

6x^2 + 12 x -7 =0

a=6
b=12
c=-7

$x= \frac{-b +/- \sqrt{b^{2} -4ac}}{2a}$

x = {-12 +/- √[12^2 - 4(6)(-7)]} / [2(6)]

x=  0.472

x = -2.472