5x^2+x-40=0

Solution for 5x^2+x-40=0 equation:

5x^2+x-40=0

a = 5; b = 1; c = -40;
Δ = b2-4ac
Δ = 12-4·5·(-40)
Δ = 801
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:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{89}}{2*5}=\frac{-1-3\sqrt{89}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{89}}{2*5}=\frac{-1+3\sqrt{89}}{10}
$