# How do you solve 2x^2 + 7 = 3x?

##### 2 Answers
Jul 18, 2015

This equation has no real solutions.

#### Explanation:

When we encounter a quadratic equation, we often want to write it in the form $a {x}^{2} + b x + c = 0$. In this case this gives $2 {x}^{2} - 3 x + 7 = 0$.

Sometimes it is easy to spot solutions, for instance is we have ${x}^{2} + 2 x + 1 = 0$, it is easy to see that ${\left(x + 1\right)}^{2} = {x}^{2} + 2 x + 1 = 0$, so it has the solution $x = - 1$, however, in this case, this is not easy to see, so we need something called the quadratic formula.

When we again look at the general form $a {x}^{2} + b x + c = 0$, we can find solutions via
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.
In this case we get $x = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 2 \cdot 7}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 56}}{4} = \frac{3}{4} \pm \frac{1}{4} \sqrt{- 47}$. Since there is no real number that satisfies this equation, because sqrt(-47) is no real number, the equation $2 {x}^{2} + 7 = 3 x$ has no real solutions. (There are however so called complex solutions, which use a constructed number $i$ such that ${i}^{2} = - 1$, but since this may be too abstract I will not go into this.)

Jul 18, 2015

Use the quadratic formula to find the complex solutions:

$x = \frac{3 \pm i \sqrt{47}}{4}$

#### Explanation:

First subtract $3 x$ from both sides to get:

$2 {x}^{2} - 3 x + 7 = 0$

This is in the form $a {x}^{2} + b x + c = 0$, with $a = 2$, $b = - 3$ and $c = 7$.

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

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - \left(4 \times 2 \times 7\right) = 9 - 56 = - 47$

Since $\Delta < 0$ the quadratic has no solutions in real numbers. It has a pair of distinct complex solutions which are complex conjugates of one another.

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{3 \pm i \sqrt{47}}{4}$