4(2x^2+43x-228)=0

Solution for 4(2x^2+43x-228)=0 equation:

4(2x^2+43x-228)=0

We multiply parentheses

8x^2+172x-912=0

a = 8; b = 172; c = -912;
Δ = b2-4ac
Δ = 1722-4·8·(-912)
Δ = 58768
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:

$\sqrt{\Delta}=\sqrt{58768}=\sqrt{16*3673}=\sqrt{16}*\sqrt{3673}=4\sqrt{3673}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(172)-4\sqrt{3673}}{2*8}=\frac{-172-4\sqrt{3673}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(172)+4\sqrt{3673}}{2*8}=\frac{-172+4\sqrt{3673}}{16}
$