35r^2+24r-35=0

Solution for 35r^2+24r-35=0 equation:

35r^2+24r-35=0

a = 35; b = 24; c = -35;
Δ = b2-4ac
Δ = 242-4·35·(-35)
Δ = 5476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5476}=74$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-74}{2*35}=\frac{-98}{70}
=-1+2/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+74}{2*35}=\frac{50}{70}
=5/7
$