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(4x-3)(2x+3)=3
We move all terms to the left:
(4x-3)(2x+3)-(3)=0
We multiply parentheses ..
(+8x^2+12x-6x-9)-3=0
We get rid of parentheses
8x^2+12x-6x-9-3=0
We add all the numbers together, and all the variables
8x^2+6x-12=0
a = 8; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·8·(-12)
Δ = 420
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{105}}{2*8}=\frac{-6-2\sqrt{105}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{105}}{2*8}=\frac{-6+2\sqrt{105}}{16} $
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