10x^2+20x-15=0

Solution for 10x^2+20x-15=0 equation:

10x^2+20x-15=0

a = 10; b = 20; c = -15;
Δ = b2-4ac
Δ = 202-4·10·(-15)
Δ = 1000
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:

$\sqrt{\Delta}=\sqrt{1000}=\sqrt{100*10}=\sqrt{100}*\sqrt{10}=10\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{10}}{2*10}=\frac{-20-10\sqrt{10}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{10}}{2*10}=\frac{-20+10\sqrt{10}}{20}
$