-28r+-4r2=0

Solution for -28r+-4r2=0 equation:

-28r+-4r^2=0

We add all the numbers together, and all the variables

-4r^2-28r=0

a = -4; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·(-4)·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*-4}=\frac{0}{-8}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*-4}=\frac{56}{-8}
=-7
$