Answer
$x_{1}=-2+\sqrt7$
$x_{2}=-2-\sqrt7$
Work Step by Step
We have to solve the quadratic equation
$$x^2+4x=3.$$
Let's rewrite it in the standard form:
$$x^2+4x-3=0.$$
Let's find the discriminant.
$$D=b^2-4ac=4^2-4\cdot1\cdot(-3)=28>0.$$
Because the discriminant is positive we have two solutions. We determine them using the Quadratic Formula:
$x_{1}=\dfrac{-b+\sqrt D}{2a}=\dfrac{-4+\sqrt{28}}{2}=-2+\dfrac{\sqrt{4\cdot7}}{2}=-2+\sqrt7$
$x_{2}=\dfrac{-b-\sqrt D}{2a}=\dfrac{-4-\sqrt{28}}{2}=-2-\dfrac{\sqrt{4\cdot7}}{2}=-2-\sqrt7$
