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5x^2+1=21
We move all terms to the left:
5x^2+1-(21)=0
We add all the numbers together, and all the variables
5x^2-20=0
a = 5; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·5·(-20)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*5}=\frac{-20}{10} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*5}=\frac{20}{10} =2 $
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