10r^2+10r-59=0

Solution for 10r^2+10r-59=0 equation:

10r^2+10r-59=0

a = 10; b = 10; c = -59;
Δ = b2-4ac
Δ = 102-4·10·(-59)
Δ = 2460
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2460}=\sqrt{4*615}=\sqrt{4}*\sqrt{615}=2\sqrt{615}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{615}}{2*10}=\frac{-10-2\sqrt{615}}{20}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{615}}{2*10}=\frac{-10+2\sqrt{615}}{20}
$