5x^2+40x+18=-47

Solution for 5x^2+40x+18=-47 equation:

5x^2+40x+18=-47

We move all terms to the left:

5x^2+40x+18-(-47)=0

We add all the numbers together, and all the variables

5x^2+40x+65=0

a = 5; b = 40; c = +65;
Δ = b2-4ac
Δ = 402-4·5·65
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{3}}{2*5}=\frac{-40-10\sqrt{3}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{3}}{2*5}=\frac{-40+10\sqrt{3}}{10}
$