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3a^2-94a-64=0
a = 3; b = -94; c = -64;
Δ = b2-4ac
Δ = -942-4·3·(-64)
Δ = 9604
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{9604}=98$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-94)-98}{2*3}=\frac{-4}{6} =-2/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-94)+98}{2*3}=\frac{192}{6} =32 $
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