(7+u)(2u+5)=0

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Solution for (7+u)(2u+5)=0 equation:



(7+u)(2u+5)=0
We add all the numbers together, and all the variables
(u+7)(2u+5)=0
We multiply parentheses ..
(+2u^2+5u+14u+35)=0
We get rid of parentheses
2u^2+5u+14u+35=0
We add all the numbers together, and all the variables
2u^2+19u+35=0
a = 2; b = 19; c = +35;
Δ = b2-4ac
Δ = 192-4·2·35
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-9}{2*2}=\frac{-28}{4} =-7 $
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+9}{2*2}=\frac{-10}{4} =-2+1/2 $

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