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3a^2+40a+100=0
a = 3; b = 40; c = +100;
Δ = b2-4ac
Δ = 402-4·3·100
Δ = 400
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{400}=20$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20}{2*3}=\frac{-60}{6} =-10 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20}{2*3}=\frac{-20}{6} =-3+1/3 $
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