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5x(3x-5)=19
We move all terms to the left:
5x(3x-5)-(19)=0
We multiply parentheses
15x^2-25x-19=0
a = 15; b = -25; c = -19;
Δ = b2-4ac
Δ = -252-4·15·(-19)
Δ = 1765
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{-(-25)-\sqrt{1765}}{2*15}=\frac{25-\sqrt{1765}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{1765}}{2*15}=\frac{25+\sqrt{1765}}{30} $
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