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7x^2+10=213
We move all terms to the left:
7x^2+10-(213)=0
We add all the numbers together, and all the variables
7x^2-203=0
a = 7; b = 0; c = -203;
Δ = b2-4ac
Δ = 02-4·7·(-203)
Δ = 5684
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5684}=\sqrt{196*29}=\sqrt{196}*\sqrt{29}=14\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{29}}{2*7}=\frac{0-14\sqrt{29}}{14} =-\frac{14\sqrt{29}}{14} =-\sqrt{29} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{29}}{2*7}=\frac{0+14\sqrt{29}}{14} =\frac{14\sqrt{29}}{14} =\sqrt{29} $
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