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(2x+5)(2x-3)=3x-12
We move all terms to the left:
(2x+5)(2x-3)-(3x-12)=0
We get rid of parentheses
(2x+5)(2x-3)-3x+12=0
We multiply parentheses ..
(+4x^2-6x+10x-15)-3x+12=0
We get rid of parentheses
4x^2-6x+10x-3x-15+12=0
We add all the numbers together, and all the variables
4x^2+x-3=0
a = 4; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·4·(-3)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*4}=\frac{-8}{8} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*4}=\frac{6}{8} =3/4 $
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