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3a^2+7a-32=0
a = 3; b = 7; c = -32;
Δ = b2-4ac
Δ = 72-4·3·(-32)
Δ = 433
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{433}}{2*3}=\frac{-7-\sqrt{433}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{433}}{2*3}=\frac{-7+\sqrt{433}}{6} $
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