4x^2-64x+228=0

Solution for 4x^2-64x+228=0 equation:

4x^2-64x+228=0

a = 4; b = -64; c = +228;
Δ = b2-4ac
Δ = -642-4·4·228
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-8\sqrt{7}}{2*4}=\frac{64-8\sqrt{7}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+8\sqrt{7}}{2*4}=\frac{64+8\sqrt{7}}{8}
$