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5x+56=(3x-6)(3x-6)
We move all terms to the left:
5x+56-((3x-6)(3x-6))=0
We multiply parentheses ..
-((+9x^2-18x-18x+36))+5x+56=0
We calculate terms in parentheses: -((+9x^2-18x-18x+36)), so:We add all the numbers together, and all the variables
(+9x^2-18x-18x+36)
We get rid of parentheses
9x^2-18x-18x+36
We add all the numbers together, and all the variables
9x^2-36x+36
Back to the equation:
-(9x^2-36x+36)
5x-(9x^2-36x+36)+56=0
We get rid of parentheses
-9x^2+5x+36x-36+56=0
We add all the numbers together, and all the variables
-9x^2+41x+20=0
a = -9; b = 41; c = +20;
Δ = b2-4ac
Δ = 412-4·(-9)·20
Δ = 2401
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2401}=49$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-49}{2*-9}=\frac{-90}{-18} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+49}{2*-9}=\frac{8}{-18} =-4/9 $
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