10x2+10x-2.5=0

Solution for 10x2+10x-2.5=0 equation:

10x^2+10x-2.5=0

a = 10; b = 10; c = -2.5;
Δ = b2-4ac
Δ = 102-4·10·(-2.5)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{2}}{2*10}=\frac{-10-10\sqrt{2}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{2}}{2*10}=\frac{-10+10\sqrt{2}}{20}
$