5r^2+2r-627=0

Solution for 5r^2+2r-627=0 equation:

5r^2+2r-627=0

a = 5; b = 2; c = -627;
Δ = b2-4ac
Δ = 22-4·5·(-627)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-112}{2*5}=\frac{-114}{10}
=-11+2/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+112}{2*5}=\frac{110}{10}
=11
$