0.01x^2+0.4x=7

Solution for 0.01x^2+0.4x=7 equation:

0.01x^2+0.4x=7

We move all terms to the left:

0.01x^2+0.4x-(7)=0

a = 0.01; b = 0.4; c = -7;
Δ = b2-4ac
Δ = 0.42-4·0.01·(-7)
Δ = 0.44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.4)-\sqrt{0.44}}{2*0.01}=\frac{-0.4-\sqrt{0.44}}{0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.44}}{2*0.01}=\frac{-0.4+\sqrt{0.44}}{0.02}
$