x2+42x+320=0

Solution for x2+42x+320=0 equation:

x2+42x+320=0

We add all the numbers together, and all the variables

x^2+42x+320=0

a = 1; b = 42; c = +320;
Δ = b2-4ac
Δ = 422-4·1·320
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-22}{2*1}=\frac{-64}{2}
=-32
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+22}{2*1}=\frac{-20}{2}
=-10
$