10x^2+10x-3=0

Solution for 10x^2+10x-3=0 equation:

10x^2+10x-3=0

a = 10; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·10·(-3)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{55}}{2*10}=\frac{-10-2\sqrt{55}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{55}}{2*10}=\frac{-10+2\sqrt{55}}{20}
$