Math Problem Solver

Solve by factoring:

Transform the equation such that one side is equal to zero. $$\begin{aligned} {x}^{2} + {\left( x + 3 \right)}^{2} &= 29 \\ 2 x^{2} + 6 x + 9 &= 29 \\ x^{2} + \left(x + 3\right)^{2} - 29 &= 0\end{aligned}$$

Factor the nonzero side of the equation completely. $$\begin{aligned} x^{2} + \left(x + 3\right)^{2} - 29 &= 0 \\ 2 \left(x - 2\right) \left(x + 5\right) &= 0\end{aligned}$$

The product of any number and zero is zero, so the equation will hold true if either of the factors is equal to zero. Set each of the factors equal to zero and solve. $$\begin{aligned} x - 2 &= 0\\ x &= 2\end{aligned}$$ $$\begin{aligned} x + 5 &= 0\\ x &= -5\end{aligned}$$

Solve by quadratic formula:

Subtract the right hand side from the left hand side of: \[{x}^{2} + {\left( x + 3 \right)}^{2} = 29\] The result is a quadratic equation: \[2 x^{2} + 6 x - 20 = 0\]

Given a quadratic equation \(a x^{2} + b x + c = 0\), where \(a\), \(b\), \(c\) are constants and \(a \ne 0\), the solutions are given by the quadratic formula: \[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\] In this case \(a=2\), \(b=6\), and \(c=-20\).

The discriminant is the quantity under the square root sign in the quadratic formula, and its sign determines the number of solutions to the quadratic equation when the coefficients are real. The discriminant is:\[b^{2}-4ac = 6^{2} - 8 \left(-20\right)=196 > 0\] The discriminant is greater than zero, so this quadratic equation has two real solutions.

The two solutions are: \[x = \frac{\left(-1\right) 6 + \sqrt{196}}{2 \cdot 2} = 2 \] \[x = \frac{\left(-1\right) 6 - \sqrt{196}}{2 \cdot 2} = -5 \]