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q^2+50=450
We move all terms to the left:
q^2+50-(450)=0
We add all the numbers together, and all the variables
q^2-400=0
a = 1; b = 0; c = -400;
Δ = b2-4ac
Δ = 02-4·1·(-400)
Δ = 1600
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{1600}=40$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*1}=\frac{-40}{2} =-20 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*1}=\frac{40}{2} =20 $
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