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x(5x-19)=4
We move all terms to the left:
x(5x-19)-(4)=0
We multiply parentheses
5x^2-19x-4=0
a = 5; b = -19; c = -4;
Δ = b2-4ac
Δ = -192-4·5·(-4)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*5}=\frac{-2}{10} =-1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*5}=\frac{40}{10} =4 $
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