\(x^{2} +6x-7=0\)
Solve by formula : \(x_{1.2} = \frac{-b+ \sqrt{ x^{2-4ac} } }{2a}\)
\(a=1,b=6,c=-7\)
\(x_{1.2} = \frac{(-6)+ \sqrt{ 6^{2}-4(-7) } }{2}\)
\(x_{1.2}= \frac{\sqrt{(-6)+ \sqrt{ 6^{2}+28 } } }{2}\)
\(x_{1.2} = \frac{(-6)+ \sqrt{36+28 } }{2}\)
\(x_{1.2} = \frac{(-6)+ \sqrt{64} }{2}\)
\(x_{1.2} = \frac{(-6)+8}{2}\)
First solution:
\(x_{1} = \frac{(-6)+8}{2}\)
\(x_{1}= \frac{
ot 2}{
ot 2}\)
\(x_{1} =1\)
Second solution:
\(x_{2} = \frac{-6-8}{2}\)
\(x_{2} =-14:2\)
\(x_{2} =-7\)
The answer is: \(x_{1} =1\) , \(x_{2} =-7\)
