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q2+14q=32
We move all terms to the left:
q2+14q-(32)=0
We add all the numbers together, and all the variables
q^2+14q-32=0
a = 1; b = 14; c = -32;
Δ = b2-4ac
Δ = 142-4·1·(-32)
Δ = 324
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{324}=18$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-18}{2*1}=\frac{-32}{2} =-16 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+18}{2*1}=\frac{4}{2} =2 $
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