r2+25r-10=0

Solution for r2+25r-10=0 equation:

r2+25r-10=0

We add all the numbers together, and all the variables

r^2+25r-10=0

a = 1; b = 25; c = -10;
Δ = b2-4ac
Δ = 252-4·1·(-10)
Δ = 665
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{665}}{2*1}=\frac{-25-\sqrt{665}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{665}}{2*1}=\frac{-25+\sqrt{665}}{2}
$