(100-10x)(3+0.5x)=300

Solution for (100-10x)(3+0.5x)=300 equation:

(100-10x)(3+0.5x)=300

We move all terms to the left:

(100-10x)(3+0.5x)-(300)=0

We add all the numbers together, and all the variables

(-10x+100)(0.5x+3)-300=0

We multiply parentheses ..

(+0x^2-30x+0x+300)-300=0

We get rid of parentheses

0x^2-30x+0x+300-300=0

We add all the numbers together, and all the variables

x^2-29x=0

a = 1; b = -29; c = 0;
Δ = b2-4ac
Δ = -292-4·1·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-29}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+29}{2*1}=\frac{58}{2}
=29
$