40x^2+50x-15=0

Solution for 40x^2+50x-15=0 equation:

40x^2+50x-15=0

a = 40; b = 50; c = -15;
Δ = b2-4ac
Δ = 502-4·40·(-15)
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-70}{2*40}=\frac{-120}{80}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+70}{2*40}=\frac{20}{80}
=1/4
$