x^2+10x-148=0

Solution for x^2+10x-148=0 equation:

x^2+10x-148=0

a = 1; b = 10; c = -148;
Δ = b2-4ac
Δ = 102-4·1·(-148)
Δ = 692
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:

$\sqrt{\Delta}=\sqrt{692}=\sqrt{4*173}=\sqrt{4}*\sqrt{173}=2\sqrt{173}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{173}}{2*1}=\frac{-10-2\sqrt{173}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{173}}{2*1}=\frac{-10+2\sqrt{173}}{2}
$