(21+x)(10+x)=26

Solution for (21+x)(10+x)=26 equation:

(21+x)(10+x)=26

We move all terms to the left:

(21+x)(10+x)-(26)=0

We add all the numbers together, and all the variables

(x+21)(x+10)-26=0

We multiply parentheses ..

(+x^2+10x+21x+210)-26=0

We get rid of parentheses

x^2+10x+21x+210-26=0

We add all the numbers together, and all the variables

x^2+31x+184=0

a = 1; b = 31; c = +184;
Δ = b2-4ac
Δ = 312-4·1·184
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-15}{2*1}=\frac{-46}{2}
=-23
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+15}{2*1}=\frac{-16}{2}
=-8
$