2,5x^2+10x=15

Solution for 2,5x^2+10x=15 equation:

2.5x^2+10x=15

We move all terms to the left:

2.5x^2+10x-(15)=0

a = 2.5; b = 10; c = -15;
Δ = b2-4ac
Δ = 102-4·2.5·(-15)
Δ = 250
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:

$\sqrt{\Delta}=\sqrt{250}=\sqrt{25*10}=\sqrt{25}*\sqrt{10}=5\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-5\sqrt{10}}{2*2.5}=\frac{-10-5\sqrt{10}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+5\sqrt{10}}{2*2.5}=\frac{-10+5\sqrt{10}}{5}
$