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9x-12=(5x+2)(2x-6)
We move all terms to the left:
9x-12-((5x+2)(2x-6))=0
We multiply parentheses ..
-((+10x^2-30x+4x-12))+9x-12=0
We calculate terms in parentheses: -((+10x^2-30x+4x-12)), so:We add all the numbers together, and all the variables
(+10x^2-30x+4x-12)
We get rid of parentheses
10x^2-30x+4x-12
We add all the numbers together, and all the variables
10x^2-26x-12
Back to the equation:
-(10x^2-26x-12)
9x-(10x^2-26x-12)-12=0
We get rid of parentheses
-10x^2+9x+26x+12-12=0
We add all the numbers together, and all the variables
-10x^2+35x=0
a = -10; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·(-10)·0
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*-10}=\frac{-70}{-20} =3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*-10}=\frac{0}{-20} =0 $
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