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2x+3-5x(2x+2)=3x+5
We move all terms to the left:
2x+3-5x(2x+2)-(3x+5)=0
We multiply parentheses
-10x^2+2x-10x-(3x+5)+3=0
We get rid of parentheses
-10x^2+2x-10x-3x-5+3=0
We add all the numbers together, and all the variables
-10x^2-11x-2=0
a = -10; b = -11; c = -2;
Δ = b2-4ac
Δ = -112-4·(-10)·(-2)
Δ = 41
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{-(-11)-\sqrt{41}}{2*-10}=\frac{11-\sqrt{41}}{-20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{41}}{2*-10}=\frac{11+\sqrt{41}}{-20} $
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