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3a^2+28a-160=0
a = 3; b = 28; c = -160;
Δ = b2-4ac
Δ = 282-4·3·(-160)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2704}=52$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-52}{2*3}=\frac{-80}{6} =-13+1/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+52}{2*3}=\frac{24}{6} =4 $
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