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4x^2-4x-5=19
We move all terms to the left:
4x^2-4x-5-(19)=0
We add all the numbers together, and all the variables
4x^2-4x-24=0
a = 4; b = -4; c = -24;
Δ = b2-4ac
Δ = -42-4·4·(-24)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*4}=\frac{-16}{8} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*4}=\frac{24}{8} =3 $
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