Verified Answer :

calculista

Answer:

The roots of the quadratic equation are

\(x=-3+3\sqrt{2}\)

\(x=-3-3\sqrt{2}\)

Step-by-step explanation:

we have

\(x^{2} +6x-9=0\)

we know that

The formula to solve a quadratic equation of the form \(ax^{2} +bx+c=0\) is equal to

\(x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}\)

in this problem we have

\(x^{2} +6x-9=0\)

so

\(a=1\\b=6\\c=-9\)

substitute in the formula

\(x=\frac{-6(+/-)\sqrt{6^{2}-4(1)(-9)}} {2(1)}\)

\(x=\frac{-6(+/-)\sqrt{72}} {2}\)

\(x=\frac{-6(+/-)6\sqrt{2}} {2}\)

\(x=-3(+/-)3\sqrt{2}\)

\(x=-3(+)3\sqrt{2}\)

\(x=-3(-)3\sqrt{2}\)

Given equation is :

\(x^{2}+6x-9=0\)

We will solve this quadratic equation, using the formula:

\(x1,2=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\) and

\(x1,2=\frac{-b-\sqrt{b^{2}-4ac}}{2a}\)

Now, putting a=1 , b=6 and c =-9 we get

\(x=\frac{-6+\sqrt{6^{2}-4*1(-9)}}{2*1}\) and

\(x=\frac{-6-\sqrt{6^{2}-4*1(-9)}}{2*1}\)

Final solutions are:  \(x=3(\sqrt{2}-1)\) and

\(x=-3(\sqrt{2}+1)\)

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