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(x+10)(2x+5)-(x+2)(x-2)=150
We move all terms to the left:
(x+10)(2x+5)-(x+2)(x-2)-(150)=0
We use the square of the difference formula
x^2+(x+10)(2x+5)+4-150=0
We multiply parentheses ..
x^2+(+2x^2+5x+20x+50)+4-150=0
We add all the numbers together, and all the variables
x^2+(+2x^2+5x+20x+50)-146=0
We get rid of parentheses
x^2+2x^2+5x+20x+50-146=0
We add all the numbers together, and all the variables
3x^2+25x-96=0
a = 3; b = 25; c = -96;
Δ = b2-4ac
Δ = 252-4·3·(-96)
Δ = 1777
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{1777}}{2*3}=\frac{-25-\sqrt{1777}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{1777}}{2*3}=\frac{-25+\sqrt{1777}}{6} $
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