2r^2+5r-25=0

Solution for 2r^2+5r-25=0 equation:

2r^2+5r-25=0

a = 2; b = 5; c = -25;
Δ = b2-4ac
Δ = 52-4·2·(-25)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-15}{2*2}=\frac{-20}{4}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+15}{2*2}=\frac{10}{4}
=2+1/2
$