40x^2+80x-70=0

Solution for 40x^2+80x-70=0 equation:

40x^2+80x-70=0

a = 40; b = 80; c = -70;
Δ = b2-4ac
Δ = 802-4·40·(-70)
Δ = 17600
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:

$\sqrt{\Delta}=\sqrt{17600}=\sqrt{1600*11}=\sqrt{1600}*\sqrt{11}=40\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-40\sqrt{11}}{2*40}=\frac{-80-40\sqrt{11}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+40\sqrt{11}}{2*40}=\frac{-80+40\sqrt{11}}{80}
$