40x-x^2=290

Solution for 40x-x^2=290 equation:

40x-x^2=290

We move all terms to the left:

40x-x^2-(290)=0

We add all the numbers together, and all the variables

-1x^2+40x-290=0

a = -1; b = 40; c = -290;
Δ = b2-4ac
Δ = 402-4·(-1)·(-290)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{110}}{2*-1}=\frac{-40-2\sqrt{110}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{110}}{2*-1}=\frac{-40+2\sqrt{110}}{-2}
$