Respuesta :
The solution to the equation is [tex]x=\frac{5-\sqrt{13}}{2}[/tex] and [tex]x=\frac{5+\sqrt{13}}{2}[/tex]
Explanation:
Given that the equation is [tex]-x^2+5x-3=0[/tex]
We need to determine the solution of the equation.
The solution of the equation can be determined using the quadratic formula.
The quadratic formula is given by
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Hence, from the equation, we have,
[tex]a=-1,\:b=5,\:c=-3[/tex]
Substituting these values in the quadratic formula, we get,
[tex]x=\frac{-5\pm \sqrt{5^2-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}[/tex]
Simplifying, we get,
[tex]x=\frac{-5\pm \sqrt{25-12}}{-2}[/tex]
Simplifying the terms within the root, we get,
[tex]x=\frac{-5\pm \sqrt{13}}{-2}[/tex]
Thus, the roots of the equation are [tex]x=\frac{-5+ \sqrt{13}}{-2}[/tex] and [tex]x=\frac{-5- \sqrt{13}}{-2}[/tex]
Taking out the negative sign, we get,
[tex]x=\frac{-(5- \sqrt{13})}{-2}[/tex] and [tex]x=\frac{-(5+ \sqrt{13})}{-2}[/tex]
Cancelling the negative sign, we get,
[tex]x=\frac{5-\sqrt{13}}{2}[/tex] and [tex]x=\frac{5+\sqrt{13}}{2}[/tex]
Thus, the solutions of the equation are [tex]x=\frac{5-\sqrt{13}}{2}[/tex] and [tex]x=\frac{5+\sqrt{13}}{2}[/tex]
