16x^2-578x+36=0

Solution for 16x^2-578x+36=0 equation:

16x^2-578x+36=0

a = 16; b = -578; c = +36;
Δ = b2-4ac
Δ = -5782-4·16·36
Δ = 331780
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:

$\sqrt{\Delta}=\sqrt{331780}=\sqrt{4*82945}=\sqrt{4}*\sqrt{82945}=2\sqrt{82945}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-578)-2\sqrt{82945}}{2*16}=\frac{578-2\sqrt{82945}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-578)+2\sqrt{82945}}{2*16}=\frac{578+2\sqrt{82945}}{32}
$