-32=2u(u-7)-5u

Solution for -32=2u(u-7)-5u equation:

-32=2u(u-7)-5u

We move all terms to the left:

-32-(2u(u-7)-5u)=0


We calculate terms in parentheses: -(2u(u-7)-5u), so:

2u(u-7)-5u

We add all the numbers together, and all the variables

-5u+2u(u-7)

We multiply parentheses

2u^2-5u-14u

We add all the numbers together, and all the variables

2u^2-19u

Back to the equation:

-(2u^2-19u)


We get rid of parentheses

-2u^2+19u-32=0

a = -2; b = 19; c = -32;
Δ = b2-4ac
Δ = 192-4·(-2)·(-32)
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{105}}{2*-2}=\frac{-19-\sqrt{105}}{-4}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{105}}{2*-2}=\frac{-19+\sqrt{105}}{-4}
$