5x^2+10x-48=0

Solution for 5x^2+10x-48=0 equation:

5x^2+10x-48=0

a = 5; b = 10; c = -48;
Δ = b2-4ac
Δ = 102-4·5·(-48)
Δ = 1060
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:

$\sqrt{\Delta}=\sqrt{1060}=\sqrt{4*265}=\sqrt{4}*\sqrt{265}=2\sqrt{265}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{265}}{2*5}=\frac{-10-2\sqrt{265}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{265}}{2*5}=\frac{-10+2\sqrt{265}}{10}
$