Solve Using the Quadratic Formula x(2x-3)=35

$x (2 x - 3) = 35$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $x (2 x - 3)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (2 x) + x \cdot - 3 = 35$

Reorder.

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot - 3 = 35$

Move $- 3$ to the left of $x$ .

$2 x \cdot x - 3 \cdot x = 35$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 (x \cdot x) - 3 \cdot x = 35$

Multiply $x$ by $x$ .

$2 x^{2} - 3 \cdot x = 35$

$2 x^{2} - 3 x = 35$

Subtract $35$ from both sides of the equation.

$2 x^{2} - 3 x - 35 = 0$

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 $a = 2$ , $b = - 3$ , and $c = - 35$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 35)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 35}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 35$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 35}}{2 \cdot 2}$

Multiply $- 8$ by $- 35$ .

$x = \frac{3 \pm \sqrt{9 + 280}}{2 \cdot 2}$

Add $9$ and $280$ .

$x = \frac{3 \pm \sqrt{289}}{2 \cdot 2}$

Rewrite $289$ as $17^{2}$ .

$x = \frac{3 \pm \sqrt{17^{2}}}{2 \cdot 2}$

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

$x = \frac{3 \pm 17}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm 17}{4}$

Step 5

The final answer is the combination of both solutions.

$x = 5, - \frac{7}{2}$

Solve Using the Quadratic Formula x(2x-3)=35

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