8u-40=-4u(u-5)

Solution for 8u-40=-4u(u-5) equation:

8u-40=-4u(u-5)

We move all terms to the left:

8u-40-(-4u(u-5))=0


We calculate terms in parentheses: -(-4u(u-5)), so:

-4u(u-5)

We multiply parentheses

-4u^2+20u

Back to the equation:

-(-4u^2+20u)


We get rid of parentheses

4u^2-20u+8u-40=0

We add all the numbers together, and all the variables

4u^2-12u-40=0

a = 4; b = -12; c = -40;
Δ = b2-4ac
Δ = -122-4·4·(-40)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-28}{2*4}=\frac{-16}{8}
=-2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+28}{2*4}=\frac{40}{8}
=5
$