2x^2-10x-58=0

Solution for 2x^2-10x-58=0 equation:

2x^2-10x-58=0

a = 2; b = -10; c = -58;
Δ = b2-4ac
Δ = -102-4·2·(-58)
Δ = 564
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:

$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{141}}{2*2}=\frac{10-2\sqrt{141}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{141}}{2*2}=\frac{10+2\sqrt{141}}{4}
$