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(2x-3)(2x-3)-16=0
We multiply parentheses ..
(+4x^2-6x-6x+9)-16=0
We get rid of parentheses
4x^2-6x-6x+9-16=0
We add all the numbers together, and all the variables
4x^2-12x-7=0
a = 4; b = -12; c = -7;
Δ = b2-4ac
Δ = -122-4·4·(-7)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-16}{2*4}=\frac{-4}{8} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+16}{2*4}=\frac{28}{8} =3+1/2 $
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