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3a^2-5a-100=0
a = 3; b = -5; c = -100;
Δ = b2-4ac
Δ = -52-4·3·(-100)
Δ = 1225
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{1225}=35$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-35}{2*3}=\frac{-30}{6} =-5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+35}{2*3}=\frac{40}{6} =6+2/3 $
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