3x^2=-16x+8

Solution for 3x^2=-16x+8 equation:

3x^2=-16x+8

We move all terms to the left:

3x^2-(-16x+8)=0

We get rid of parentheses

3x^2+16x-8=0

a = 3; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·3·(-8)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{22}}{2*3}=\frac{-16-4\sqrt{22}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{22}}{2*3}=\frac{-16+4\sqrt{22}}{6}
$