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3-14x-5x^2=0
a = -5; b = -14; c = +3;
Δ = b2-4ac
Δ = -142-4·(-5)·3
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-16}{2*-5}=\frac{-2}{-10} =1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+16}{2*-5}=\frac{30}{-10} =-3 $
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