5x(10x+5)=42x^2

Solution for 5x(10x+5)=42x^2 equation:

5x(10x+5)=42x^2

We move all terms to the left:

5x(10x+5)-(42x^2)=0

determiningTheFunctionDomain
-42x^2+5x(10x+5)=0

We multiply parentheses

-42x^2+50x^2+25x=0

We add all the numbers together, and all the variables

8x^2+25x=0

a = 8; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·8·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*8}=\frac{-50}{16}
=-3+1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*8}=\frac{0}{16}
=0
$