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Sagot :
Given:
[tex]y=-x^2+8x-13[/tex]To determine the amount and type of solutions, apply the quadratic formula below:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Replace 0 with y and solve for x.
Where:
a= -1
b= 8
c = -13
Thus, we have:
[tex]0=-x^2+8x-13[/tex][tex]\begin{gathered} x=\frac{-8\pm\sqrt[]{8^2-4(-1)(-13)}}{2(-1)} \\ \\ x=\frac{-8\pm\sqrt[]{64-52}}{2(-1)} \\ \\ x=\frac{-8\pm\sqrt[]{12}}{-2} \\ \\ x=\frac{-8\pm\sqrt[]{4\ast3}}{-2} \\ \\ x=\frac{-8\pm\sqrt[]{2^2\ast3}}{-2} \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} x=\frac{-8\pm2\sqrt[]{3}}{-2} \\ \\ x=4\pm2\sqrt[]{3} \\ \\ \\ x=4+2\sqrt[]{3},\text{ and 4-2}\sqrt[]{3} \end{gathered}[/tex]ANSWER:
[tex]x=4\pm2\sqrt[]{3}[/tex]
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