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3x^2-7x-40=0
a = 3; b = -7; c = -40;
Δ = b2-4ac
Δ = -72-4·3·(-40)
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-23}{2*3}=\frac{-16}{6} =-2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+23}{2*3}=\frac{30}{6} =5 $
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