7r^2+4r-15=0

Solution for 7r^2+4r-15=0 equation:

7r^2+4r-15=0

a = 7; b = 4; c = -15;
Δ = b2-4ac
Δ = 42-4·7·(-15)
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{109}}{2*7}=\frac{-4-2\sqrt{109}}{14}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{109}}{2*7}=\frac{-4+2\sqrt{109}}{14}
$