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(4x-3)(x+3)=-8
We move all terms to the left:
(4x-3)(x+3)-(-8)=0
We add all the numbers together, and all the variables
(4x-3)(x+3)+8=0
We multiply parentheses ..
(+4x^2+12x-3x-9)+8=0
We get rid of parentheses
4x^2+12x-3x-9+8=0
We add all the numbers together, and all the variables
4x^2+9x-1=0
a = 4; b = 9; c = -1;
Δ = b2-4ac
Δ = 92-4·4·(-1)
Δ = 97
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{97}}{2*4}=\frac{-9-\sqrt{97}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{97}}{2*4}=\frac{-9+\sqrt{97}}{8} $
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