(60x-0.4x2)-(3x+13)=100

Solution for (60x-0.4x2)-(3x+13)=100 equation:

(60x-0.4x^2)-(3x+13)=100

We move all terms to the left:

(60x-0.4x^2)-(3x+13)-(100)=0

We get rid of parentheses

-0.4x^2+60x-3x-13-100=0

We add all the numbers together, and all the variables

-0.4x^2+57x-113=0

a = -0.4; b = 57; c = -113;
Δ = b2-4ac
Δ = 572-4·(-0.4)·(-113)
Δ = 3068.2
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{-(57)-\sqrt{3068.2}}{2*-0.4}=\frac{-57-\sqrt{3068.2}}{-0.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+\sqrt{3068.2}}{2*-0.4}=\frac{-57+\sqrt{3068.2}}{-0.8}
$