4x-3x^2+8=0

Solution for 4x-3x^2+8=0 equation:

4x-3x^2+8=0

a = -3; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-3)·8
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{7}}{2*-3}=\frac{-4-4\sqrt{7}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{7}}{2*-3}=\frac{-4+4\sqrt{7}}{-6}
$