(2u-5)(4+u)=0

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Solution for (2u-5)(4+u)=0 equation:



(2u-5)(4+u)=0
We add all the numbers together, and all the variables
(2u-5)(u+4)=0
We multiply parentheses ..
(+2u^2+8u-5u-20)=0
We get rid of parentheses
2u^2+8u-5u-20=0
We add all the numbers together, and all the variables
2u^2+3u-20=0
a = 2; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·2·(-20)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*2}=\frac{-16}{4} =-4 $
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*2}=\frac{10}{4} =2+1/2 $

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