(51+x)(19-x)=0

Solution for (51+x)(19-x)=0 equation:

(51+x)(19-x)=0

We add all the numbers together, and all the variables

(x+51)(-1x+19)=0

We multiply parentheses ..

(-1x^2+19x-51x+969)=0

We get rid of parentheses

-1x^2+19x-51x+969=0

We add all the numbers together, and all the variables

-1x^2-32x+969=0

a = -1; b = -32; c = +969;
Δ = b2-4ac
Δ = -322-4·(-1)·969
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-70}{2*-1}=\frac{-38}{-2}
=+19
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+70}{2*-1}=\frac{102}{-2}
=-51
$