2x^2-58x+420=0

Solution for 2x^2-58x+420=0 equation:

2x^2-58x+420=0

a = 2; b = -58; c = +420;
Δ = b2-4ac
Δ = -582-4·2·420
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-58)-2}{2*2}=\frac{56}{4}
=14
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-58)+2}{2*2}=\frac{60}{4}
=15
$