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4x^2+10=206
We move all terms to the left:
4x^2+10-(206)=0
We add all the numbers together, and all the variables
4x^2-196=0
a = 4; b = 0; c = -196;
Δ = b2-4ac
Δ = 02-4·4·(-196)
Δ = 3136
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{3136}=56$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-56}{2*4}=\frac{-56}{8} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+56}{2*4}=\frac{56}{8} =7 $
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