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4x^2=-19x+5
We move all terms to the left:
4x^2-(-19x+5)=0
We get rid of parentheses
4x^2+19x-5=0
a = 4; b = 19; c = -5;
Δ = b2-4ac
Δ = 192-4·4·(-5)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*4}=\frac{-40}{8} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*4}=\frac{2}{8} =1/4 $
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