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(x-2)(4x+1)-(2x+3)=8
We move all terms to the left:
(x-2)(4x+1)-(2x+3)-(8)=0
We get rid of parentheses
(x-2)(4x+1)-2x-3-8=0
We multiply parentheses ..
(+4x^2+x-8x-2)-2x-3-8=0
We add all the numbers together, and all the variables
(+4x^2+x-8x-2)-2x-11=0
We get rid of parentheses
4x^2+x-8x-2x-2-11=0
We add all the numbers together, and all the variables
4x^2-9x-13=0
a = 4; b = -9; c = -13;
Δ = b2-4ac
Δ = -92-4·4·(-13)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-17}{2*4}=\frac{-8}{8} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+17}{2*4}=\frac{26}{8} =3+1/4 $
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