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7x^2+2x+1=014
We move all terms to the left:
7x^2+2x+1-(014)=0
We add all the numbers together, and all the variables
7x^2+2x-13=0
a = 7; b = 2; c = -13;
Δ = b2-4ac
Δ = 22-4·7·(-13)
Δ = 368
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{23}}{2*7}=\frac{-2-4\sqrt{23}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{23}}{2*7}=\frac{-2+4\sqrt{23}}{14} $
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