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7(n^2-70n+25)=0
We multiply parentheses
7n^2-490n+175=0
a = 7; b = -490; c = +175;
Δ = b2-4ac
Δ = -4902-4·7·175
Δ = 235200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{235200}=\sqrt{78400*3}=\sqrt{78400}*\sqrt{3}=280\sqrt{3}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-490)-280\sqrt{3}}{2*7}=\frac{490-280\sqrt{3}}{14} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-490)+280\sqrt{3}}{2*7}=\frac{490+280\sqrt{3}}{14} $
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