21x2+36x-15=0

Solution for 21x2+36x-15=0 equation:

21x^2+36x-15=0

a = 21; b = 36; c = -15;
Δ = b2-4ac
Δ = 362-4·21·(-15)
Δ = 2556
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2556}=\sqrt{36*71}=\sqrt{36}*\sqrt{71}=6\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{71}}{2*21}=\frac{-36-6\sqrt{71}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{71}}{2*21}=\frac{-36+6\sqrt{71}}{42}
$