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3a^2-34a+95=0
a = 3; b = -34; c = +95;
Δ = b2-4ac
Δ = -342-4·3·95
Δ = 16
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{16}=4$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-4}{2*3}=\frac{30}{6} =5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+4}{2*3}=\frac{38}{6} =6+1/3 $
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