3x^2+6x-8=5

Solution for 3x^2+6x-8=5 equation:

3x^2+6x-8=5

We move all terms to the left:

3x^2+6x-8-(5)=0

We add all the numbers together, and all the variables

3x^2+6x-13=0

a = 3; b = 6; c = -13;
Δ = b2-4ac
Δ = 62-4·3·(-13)
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-8\sqrt{3}}{2*3}=\frac{-6-8\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+8\sqrt{3}}{2*3}=\frac{-6+8\sqrt{3}}{6}
$