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30x^2-19x-5=0
a = 30; b = -19; c = -5;
Δ = b2-4ac
Δ = -192-4·30·(-5)
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-31}{2*30}=\frac{-12}{60} =-1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+31}{2*30}=\frac{50}{60} =5/6 $
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