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(2x+10)(x+5)=2344
We move all terms to the left:
(2x+10)(x+5)-(2344)=0
We multiply parentheses ..
(+2x^2+10x+10x+50)-2344=0
We get rid of parentheses
2x^2+10x+10x+50-2344=0
We add all the numbers together, and all the variables
2x^2+20x-2294=0
a = 2; b = 20; c = -2294;
Δ = b2-4ac
Δ = 202-4·2·(-2294)
Δ = 18752
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{18752}=\sqrt{64*293}=\sqrt{64}*\sqrt{293}=8\sqrt{293}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{293}}{2*2}=\frac{-20-8\sqrt{293}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{293}}{2*2}=\frac{-20+8\sqrt{293}}{4} $
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