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4x^2+4x-19.25=0
a = 4; b = 4; c = -19.25;
Δ = b2-4ac
Δ = 42-4·4·(-19.25)
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-18}{2*4}=\frac{-22}{8} =-2+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+18}{2*4}=\frac{14}{8} =1+3/4 $
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