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7x^2+26x+4=0
a = 7; b = 26; c = +4;
Δ = b2-4ac
Δ = 262-4·7·4
Δ = 564
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{141}}{2*7}=\frac{-26-2\sqrt{141}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{141}}{2*7}=\frac{-26+2\sqrt{141}}{14} $
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