x^2-62x+952=0

Solution for x^2-62x+952=0 equation:

x^2-62x+952=0

a = 1; b = -62; c = +952;
Δ = b2-4ac
Δ = -622-4·1·952
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-6}{2*1}=\frac{56}{2}
=28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+6}{2*1}=\frac{68}{2}
=34
$