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3 years ago

Whats the problem help me

Mathematics

1 answer:

rusak2 [61]3 years ago

3 0

5*23*2
5*23=115
115*2=230

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2 years ago

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The solution to the equation is $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Explanation:

Given that the equation is $-x^2+5x-3=0$

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

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Hence, from the equation, we have,

$a=-1,\:b=5,\:c=-3$

Substituting these values in the quadratic formula, we get,

$x=\frac{-5\pm \sqrt{5^2-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$

Simplifying, we get,

$x=\frac{-5\pm \sqrt{25-12}}{-2}$

Simplifying the terms within the root, we get,

$x=\frac{-5\pm \sqrt{13}}{-2}$

Thus, the roots of the equation are $x=\frac{-5+ \sqrt{13}}{-2}$ and $x=\frac{-5- \sqrt{13}}{-2}$

Taking out the negative sign, we get,

$x=\frac{-(5- \sqrt{13})}{-2}$ and $x=\frac{-(5+ \sqrt{13})}{-2}$

Cancelling the negative sign, we get,

$x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Thus, the solutions of the equation are $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

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3 years ago