# How do you use the quadratic formula to find both solutions to the quadratic equation x^2-3x=-10?

Jun 28, 2015

Solve $y = {x}^{2} - 3 x + 10 = 0$

#### Explanation:

$D = {d}^{2} = {b}^{2} - 4 a c = 9 - 40 = - 31 < 0$ --> $d = \pm i \sqrt{31}$

There are no real roots. There are 2 complex roots.

$x = \frac{3}{2} \pm \frac{i \sqrt{31}}{2}$

Jun 28, 2015

${x}^{2} - 3 x = - 10$ has only complex roots $x = \frac{3 \pm i \sqrt{31}}{2}$

${x}^{2} - 3 x = 10$ has roots $x = 5$ and $x = - 2$

#### Explanation:

I'm not sure the sign of $10$ on the right hand side is correct, so let's deal with both cases:

Case 1 ${x}^{2} - 3 x = - 10$

Add $10$ to both sides to get ${x}^{2} - 3 x + 10 = 0$

Let $f \left(x\right) = {x}^{2} - 3 x + 10$.

This is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 3$ and $c = 10$.

The discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - \left(4 \times 1 \times 10\right) = 9 - 40 = - 31$

Since this is negative $f \left(x\right) = 0$ has two distinct complex roots, given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{3 \pm i \sqrt{31}}{2}$

Case 2 ${x}^{2} - 3 x = 10$

Subtract $10$ from both sides to get ${x}^{2} - 3 x - 10 = 0$

Let $f \left(x\right) = {x}^{2} - 3 x - 10$

This is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 3$ and $c = - 10$

The discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - \left(4 \times 1 \times - 10\right) = 9 + 40 = 49 = {7}^{2}$

This is positive and a perfect square, so the roots of $f \left(x\right) = 0$ are distinct rational real numbers, given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{3 \pm 7}{2}$

That is $x = - \frac{4}{2} = - 2$ and $x = \frac{10}{2} = 5$