40x^2+22x-90=0

Solution for 40x^2+22x-90=0 equation:

40x^2+22x-90=0

a = 40; b = 22; c = -90;
Δ = b2-4ac
Δ = 222-4·40·(-90)
Δ = 14884
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14884}=122$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-122}{2*40}=\frac{-144}{80}
=-1+4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+122}{2*40}=\frac{100}{80}
=1+1/4
$