(21+x)(10+x)=350

Solution for (21+x)(10+x)=350 equation:

(21+x)(10+x)=350

We move all terms to the left:

(21+x)(10+x)-(350)=0

We add all the numbers together, and all the variables

(x+21)(x+10)-350=0

We multiply parentheses ..

(+x^2+10x+21x+210)-350=0

We get rid of parentheses

x^2+10x+21x+210-350=0

We add all the numbers together, and all the variables

x^2+31x-140=0

a = 1; b = 31; c = -140;
Δ = b2-4ac
Δ = 312-4·1·(-140)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-39}{2*1}=\frac{-70}{2}
=-35
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+39}{2*1}=\frac{8}{2}
=4
$