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7x(^2)+20=252
We move all terms to the left:
7x(^2)+20-(252)=0
We add all the numbers together, and all the variables
7x^2-232=0
a = 7; b = 0; c = -232;
Δ = b2-4ac
Δ = 02-4·7·(-232)
Δ = 6496
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6496}=\sqrt{16*406}=\sqrt{16}*\sqrt{406}=4\sqrt{406}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{406}}{2*7}=\frac{0-4\sqrt{406}}{14} =-\frac{4\sqrt{406}}{14} =-\frac{2\sqrt{406}}{7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{406}}{2*7}=\frac{0+4\sqrt{406}}{14} =\frac{4\sqrt{406}}{14} =\frac{2\sqrt{406}}{7} $
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