25r2+65r-140=0

Solution for 25r2+65r-140=0 equation:

25r^2+65r-140=0

a = 25; b = 65; c = -140;
Δ = b2-4ac
Δ = 652-4·25·(-140)
Δ = 18225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{18225}=135$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(65)-135}{2*25}=\frac{-200}{50}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(65)+135}{2*25}=\frac{70}{50}
=1+2/5
$