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

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Algebra

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

Subtract $7$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

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Simplify the numerator.

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Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 7}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 7}}{2 \cdot 2}$

Multiply $- 8$ by $- 7$ .

$x = \frac{3 \pm \sqrt{9 + 56}}{2 \cdot 2}$

Add $9$ and $56$ .

$x = \frac{3 \pm \sqrt{65}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm \sqrt{65}}{4}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3 \pm \sqrt{65}}{4}$

Decimal Form:

$x = 2.76556443 \dots, - 1.26556443 \dots$

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