Solve Using the Quadratic Formula 2x^2-7x=-3

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

Step 1

Add $3$ to both sides of the equation.

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

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 2$ , $b = - 7$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (2 \cdot 3)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 3$ .

Multiply $- 4$ by $2$ .

$x = \frac{7 \pm \sqrt{49 - 8 \cdot 3}}{2 \cdot 2}$

Multiply $- 8$ by $3$ .

$x = \frac{7 \pm \sqrt{49 - 24}}{2 \cdot 2}$

Subtract $24$ from $49$ .

$x = \frac{7 \pm \sqrt{25}}{2 \cdot 2}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{7 \pm \sqrt{5^{2}}}{2 \cdot 2}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{7 \pm 5}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{7 \pm 5}{4}$

Step 5

The final answer is the combination of both solutions.

$x = 3, \frac{1}{2}$