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6q^2+37q=0
a = 6; b = 37; c = 0;
Δ = b2-4ac
Δ = 372-4·6·0
Δ = 1369
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{1369}=37$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-37}{2*6}=\frac{-74}{12} =-6+1/6 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+37}{2*6}=\frac{0}{12} =0 $
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