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q2+16=20
We move all terms to the left:
q2+16-(20)=0
We add all the numbers together, and all the variables
q^2-4=0
a = 1; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·1·(-4)
Δ = 16
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{16}=4$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*1}=\frac{-4}{2} =-2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*1}=\frac{4}{2} =2 $
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