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(x+4)(x-6)=(2x+5)(x-6)
We move all terms to the left:
(x+4)(x-6)-((2x+5)(x-6))=0
We multiply parentheses ..
(+x^2-6x+4x-24)-((2x+5)(x-6))=0
We calculate terms in parentheses: -((2x+5)(x-6)), so:We get rid of parentheses
(2x+5)(x-6)
We multiply parentheses ..
(+2x^2-12x+5x-30)
We get rid of parentheses
2x^2-12x+5x-30
We add all the numbers together, and all the variables
2x^2-7x-30
Back to the equation:
-(2x^2-7x-30)
x^2-2x^2-6x+4x+7x-24+30=0
We add all the numbers together, and all the variables
-1x^2+5x+6=0
a = -1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·(-1)·6
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-1}=\frac{-12}{-2} =+6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-1}=\frac{2}{-2} =-1 $
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