Use the distributive property to multiply $4x$ by $x+7$.
$$4x^{2}+28x+10=2\left(x^{2}+5\right)$$
Use the distributive property to multiply $2$ by $x^{2}+5$.
$$4x^{2}+28x+10=2x^{2}+10$$
Subtract $2x^{2}$ from both sides.
$$4x^{2}+28x+10-2x^{2}=10$$
Combine $4x^{2}$ and $-2x^{2}$ to get $2x^{2}$.
$$2x^{2}+28x+10=10$$
Subtract $10$ from both sides.
$$2x^{2}+28x+10-10=0$$
Subtract $10$ from $10$ to get $0$.
$$2x^{2}+28x=0$$
Factor out $x$.
$$x\left(2x+28\right)=0$$
To find equation solutions, solve $x=0$ and $2x+28=0$.
$$x=0$$ $$x=-14$$
Steps Using the Quadratic Formula
Use the distributive property to multiply $4x$ by $x+7$.
$$4x^{2}+28x+10=2\left(x^{2}+5\right)$$
Use the distributive property to multiply $2$ by $x^{2}+5$.
$$4x^{2}+28x+10=2x^{2}+10$$
Subtract $2x^{2}$ from both sides.
$$4x^{2}+28x+10-2x^{2}=10$$
Combine $4x^{2}$ and $-2x^{2}$ to get $2x^{2}$.
$$2x^{2}+28x+10=10$$
Subtract $10$ from both sides.
$$2x^{2}+28x+10-10=0$$
Subtract $10$ from $10$ to get $0$.
$$2x^{2}+28x=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $28$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-28±\sqrt{28^{2}}}{2\times 2}$$
Take the square root of $28^{2}$.
$$x=\frac{-28±28}{2\times 2}$$
Multiply $2$ times $2$.
$$x=\frac{-28±28}{4}$$
Now solve the equation $x=\frac{-28±28}{4}$ when $±$ is plus. Add $-28$ to $28$.
$$x=\frac{0}{4}$$
Divide $0$ by $4$.
$$x=0$$
Now solve the equation $x=\frac{-28±28}{4}$ when $±$ is minus. Subtract $28$ from $-28$.
$$x=-\frac{56}{4}$$
Divide $-56$ by $4$.
$$x=-14$$
The equation is now solved.
$$x=0$$ $$x=-14$$
Steps for Completing the Square
Use the distributive property to multiply $4x$ by $x+7$.
$$4x^{2}+28x+10=2\left(x^{2}+5\right)$$
Use the distributive property to multiply $2$ by $x^{2}+5$.
$$4x^{2}+28x+10=2x^{2}+10$$
Subtract $2x^{2}$ from both sides.
$$4x^{2}+28x+10-2x^{2}=10$$
Combine $4x^{2}$ and $-2x^{2}$ to get $2x^{2}$.
$$2x^{2}+28x+10=10$$
Subtract $10$ from both sides.
$$2x^{2}+28x=10-10$$
Subtract $10$ from $10$ to get $0$.
$$2x^{2}+28x=0$$
Divide both sides by $2$.
$$\frac{2x^{2}+28x}{2}=\frac{0}{2}$$
Dividing by $2$ undoes the multiplication by $2$.
$$x^{2}+\frac{28}{2}x=\frac{0}{2}$$
Divide $28$ by $2$.
$$x^{2}+14x=\frac{0}{2}$$
Divide $0$ by $2$.
$$x^{2}+14x=0$$
Divide $14$, the coefficient of the $x$ term, by $2$ to get $7$. Then add the square of $7$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}+14x+7^{2}=7^{2}$$
Square $7$.
$$x^{2}+14x+49=49$$
Factor $x^{2}+14x+49$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+7\right)^{2}=49$$
Take the square root of both sides of the equation.
