(0.01x^2+1)(42)-(42x)(0.02x)=0

Solution for (0.01x^2+1)(42)-(42x)(0.02x)=0 equation:

(0.01x^2+1)(42)-(42x)(0.02x)=0

We add all the numbers together, and all the variables

(0.01x^2+1)42-42x(+0.02x)=0

We multiply parentheses

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

We add all the numbers together, and all the variables

-1x^2+x+42=0

a = -1; b = 1; c = +42;
Δ = b2-4ac
Δ = 12-4·(-1)·42
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*-1}=\frac{-14}{-2}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*-1}=\frac{12}{-2}
=-6
$