21x^2+90x-40=0

Solution for 21x^2+90x-40=0 equation:

21x^2+90x-40=0

a = 21; b = 90; c = -40;
Δ = b2-4ac
Δ = 902-4·21·(-40)
Δ = 11460
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:

$\sqrt{\Delta}=\sqrt{11460}=\sqrt{4*2865}=\sqrt{4}*\sqrt{2865}=2\sqrt{2865}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-2\sqrt{2865}}{2*21}=\frac{-90-2\sqrt{2865}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+2\sqrt{2865}}{2*21}=\frac{-90+2\sqrt{2865}}{42}
$