Find the Roots (Zeros) y=4x^2-3+4x

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

Step 1

Set $4 x^{2} - 3 + 4 x$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor by grouping.

Reorder terms.

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

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot - 3 = - 12$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

$4 x^{2} + 4 (x) - 3 = 0$

Rewrite $4$ as $- 2$ plus $6$

$4 x^{2} + (- 2 + 6) x - 3 = 0$

Apply the distributive property.

$4 x^{2} - 2 x + 6 x - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 x^{2} - 2 x) + 6 x - 3 = 0$

Factor out the greatest common factor (GCF) from each group.

$2 x (2 x - 1) + 3 (2 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (2 x + 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x - 1 = 0$

$2 x + 3 = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

The final solution is all the values that make $(2 x - 1) (2 x + 3) = 0$ true.

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

Step 3

Find the Roots (Zeros) y=4x^2-3+4x

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