(4u-5)(4+u)=u

Solution for (4u-5)(4+u)=u equation:

(4u-5)(4+u)=u

We move all terms to the left:

(4u-5)(4+u)-(u)=0

We add all the numbers together, and all the variables

(4u-5)(u+4)-u=0

We add all the numbers together, and all the variables

-1u+(4u-5)(u+4)=0

We multiply parentheses ..

(+4u^2+16u-5u-20)-1u=0

We get rid of parentheses

4u^2+16u-5u-1u-20=0

We add all the numbers together, and all the variables

4u^2+10u-20=0

a = 4; b = 10; c = -20;
Δ = b2-4ac
Δ = 102-4·4·(-20)
Δ = 420
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{105}}{2*4}=\frac{-10-2\sqrt{105}}{8}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{105}}{2*4}=\frac{-10+2\sqrt{105}}{8}
$