Solve Using the Quadratic Formula x(7x-3)=10

$x (7 x - 3) = 10$

Step 1

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

Simplify the left side.

Simplify $x (7 x - 3)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (7 x) + x \cdot - 3 = 10$

Reorder.

Rewrite using the commutative property of multiplication.

$7 x \cdot x + x \cdot - 3 = 10$

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

$7 x \cdot x - 3 \cdot x = 10$

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

Move $x$ .

$7 (x \cdot x) - 3 \cdot x = 10$

Multiply $x$ by $x$ .

$7 x^{2} - 3 \cdot x = 10$

$7 x^{2} - 3 x = 10$

Subtract $10$ from both sides of the equation.

$7 x^{2} - 3 x - 10 = 0$

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 $a = 7$ , $b = - 3$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (7 \cdot - 10)}}{2 \cdot 7}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 7 \cdot - 10}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot - 10$ .

Multiply $- 4$ by $7$ .

$x = \frac{3 \pm \sqrt{9 - 28 \cdot - 10}}{2 \cdot 7}$

Multiply $- 28$ by $- 10$ .

$x = \frac{3 \pm \sqrt{9 + 280}}{2 \cdot 7}$

Add $9$ and $280$ .

$x = \frac{3 \pm \sqrt{289}}{2 \cdot 7}$

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

$x = \frac{3 \pm \sqrt{17^{2}}}{2 \cdot 7}$

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

$x = \frac{3 \pm 17}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{3 \pm 17}{14}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{10}{7}, - 1$

Solve Using the Quadratic Formula x(7x-3)=10

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