45x^2+22x-80=0

Solution for 45x^2+22x-80=0 equation:

45x^2+22x-80=0

a = 45; b = 22; c = -80;
Δ = b2-4ac
Δ = 222-4·45·(-80)
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*45}=\frac{-144}{90}
=-1+3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+122}{2*45}=\frac{100}{90}
=1+1/9
$