4x^2+26x+40=200

Solution for 4x^2+26x+40=200 equation:

4x^2+26x+40=200

We move all terms to the left:

4x^2+26x+40-(200)=0

We add all the numbers together, and all the variables

4x^2+26x-160=0

a = 4; b = 26; c = -160;
Δ = b2-4ac
Δ = 262-4·4·(-160)
Δ = 3236
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:

$\sqrt{\Delta}=\sqrt{3236}=\sqrt{4*809}=\sqrt{4}*\sqrt{809}=2\sqrt{809}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{809}}{2*4}=\frac{-26-2\sqrt{809}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{809}}{2*4}=\frac{-26+2\sqrt{809}}{8}
$