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10q^2+11q-6=0
a = 10; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·10·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*10}=\frac{-30}{20} =-1+1/2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*10}=\frac{8}{20} =2/5 $
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