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3x^2+50x-408=0
a = 3; b = 50; c = -408;
Δ = b2-4ac
Δ = 502-4·3·(-408)
Δ = 7396
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{7396}=86$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-86}{2*3}=\frac{-136}{6} =-22+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+86}{2*3}=\frac{36}{6} =6 $
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