Solve Using the Quadratic Formula 4x(x-1)-5x^2=3

$4 x (x - 1) - 5 x^{2} = 3$

Step 1

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

Simplify the left side.

Simplify $4 x (x - 1) - 5 x^{2}$ .

Simplify each term.

Apply the distributive property.

$4 x \cdot x + 4 x \cdot - 1 - 5 x^{2} = 3$

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

Move $x$ .

$4 (x \cdot x) + 4 x \cdot - 1 - 5 x^{2} = 3$

Multiply $x$ by $x$ .

$4 x^{2} + 4 x \cdot - 1 - 5 x^{2} = 3$

Multiply $- 1$ by $4$ .

$4 x^{2} - 4 x - 5 x^{2} = 3$

Subtract $5 x^{2}$ from $4 x^{2}$ .

$- x^{2} - 4 x = 3$

Subtract $3$ from both sides of the equation.

$- x^{2} - 4 x - 3 = 0$

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 $a = - 1$ , $b = - 4$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (- 1 \cdot - 3)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 1 \cdot - 3}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot - 3$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{4 \pm \sqrt{16 + 4 \cdot - 3}}{2 \cdot - 1}$

Multiply $4$ by $- 3$ .

$x = \frac{4 \pm \sqrt{16 - 12}}{2 \cdot - 1}$

Subtract $12$ from $16$ .

$x = \frac{4 \pm \sqrt{4}}{2 \cdot - 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{4 \pm \sqrt{2^{2}}}{2 \cdot - 1}$

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

$x = \frac{4 \pm 2}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{4 \pm 2}{- 2}$

Simplify $\frac{4 \pm 2}{- 2}$ .

$x = \frac{2 \pm 1}{- 1}$

Move the negative one from the denominator of $\frac{2 \pm 1}{- 1}$ .

$x = - 1 \cdot (2 \pm 1)$

Rewrite $- 1 \cdot (2 \pm 1)$ as $- (2 \pm 1)$ .

$x = - (2 \pm 1)$

Step 5

The final answer is the combination of both solutions.

$x = - 3, - 1$

Solve Using the Quadratic Formula 4x(x-1)-5x^2=3

Related Questions

Support LCMGCF.com by disabling your adblocker.