3x^2+6x-7=17

Solution for 3x^2+6x-7=17 equation:

3x^2+6x-7=17

We move all terms to the left:

3x^2+6x-7-(17)=0

We add all the numbers together, and all the variables

3x^2+6x-24=0

a = 3; b = 6; c = -24;
Δ = b2-4ac
Δ = 62-4·3·(-24)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*3}=\frac{12}{6}
=2
$