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To find the roots of a quadratic equation, plug its coefficients ($$a$$, $$b$$ and $$c$$ ) into the quadratic formula:

$$x=\left(-b\pm sqrt\left(b^2-4ac\right)\right)/\left(2a\right)$$

$$x=\left(-10\pm sqrt\left({10}^2-4*-1*-25\right)\right)/\left(2*-1\right)$$

$$x=\left(-10\pm sqrt\left(100-4*-1*-25\right)\right)/\left(2*-1\right)$$

$$x=\left(-10\pm sqrt\left(100--4*-25\right)\right)/\left(2*-1\right)$$

$$x=\left(-10\pm sqrt\left(100-100\right)\right)/\left(2*-1\right)$$

$$x=\left(-10\pm sqrt\left(0\right)\right)/\left(2*-1\right)$$

$$x=\left(-10\pm sqrt\left(0\right)\right)/\left(-2\right)$$

to get the result:

$$x=\left(-10\pm sqrt\left(0\right)\right)/\left(-2\right)$$

Simplify $$0$$ by finding its prime factors:

The prime factorization of $$0$$ is $$0$$

$$x=\left(-10\pm 0\right)/\left(-2\right)$$

The ± means two roots are possible, but since zero is result of the square root, we have one root:

Separate the equations:
$$x_1=\left(-10+0\right)/\left(-2\right)$$ and $$x_2=\left(-10-0\right)/\left(-2\right)$$

$$x_1=\left(-10+0\right)/\left(-2\right)$$

$$x_1=\left(-10+0\right)/\left(-2\right)$$

$$x_1=\left(-10\right)/\left(-2\right)$$

$$x_1=-\frac{10}{-}2$$

$$x_1=5$$

To find the intervals of a quadratic inequality, we start by finding its parabola.

The roots of the parabola (where it meets the x-axis) are: 5.

Since the $$a$$ coefficient is negative ($$a$$=-1), this is a "negative" quadratic inequality and the parabola points downward, like a frown.

If the inequality sign is ≤ or ≥ , then the intervals include the roots and we use a solid line. If the inequality sign is < or > the intervals do not include the roots and we use a dotted line.

Since $$-1x^2+10x-25>0$$ has a $$>$$ inequality sign, we look for the parabola intervals that are above the x-axis.

Solution: $$$$

Interval notation: $$$$