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(4x-22)+(10x-4)(x+11)=18
We move all terms to the left:
(4x-22)+(10x-4)(x+11)-(18)=0
We get rid of parentheses
4x+(10x-4)(x+11)-22-18=0
We multiply parentheses ..
(+10x^2+110x-4x-44)+4x-22-18=0
We add all the numbers together, and all the variables
(+10x^2+110x-4x-44)+4x-40=0
We get rid of parentheses
10x^2+110x-4x+4x-44-40=0
We add all the numbers together, and all the variables
10x^2+110x-84=0
a = 10; b = 110; c = -84;
Δ = b2-4ac
Δ = 1102-4·10·(-84)
Δ = 15460
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{15460}=\sqrt{4*3865}=\sqrt{4}*\sqrt{3865}=2\sqrt{3865}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(110)-2\sqrt{3865}}{2*10}=\frac{-110-2\sqrt{3865}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(110)+2\sqrt{3865}}{2*10}=\frac{-110+2\sqrt{3865}}{20} $
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