Answer:
![x = 0, -(5)/(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/5g1eqa1p8008y5w0eey8ruq9sjzm9uvnr2.png)
Explanation:
First move all terms to one side of the equation to set them equal to 0
That is already done in your problem
The try to factor, if possible. In your case, factor out x
x(2x + 5) = 0 Now, use the property that says if AB = 0, then A = 0 or B = 0 or both A and B equal 0
So, x = 0 or 2x + 5 = 0
2x = -5
x =
![-(5)/(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/orgi5pum3e3fnacwg3rbfv6auqx11x6m6n.png)
Always check your results in the original equation to see if they work
![2(0^(2) + 5(0) = 0 + 0 = 0\\](https://img.qammunity.org/2022/formulas/mathematics/high-school/vnsljsb982pirgqmosbsadde0vhiieu8d0.png)
and
![2(-(5)/(2) )^(2) + 5(-(5)/(2) )\\ = 2((25)/(4) ) - (25)/(2) \\ = (25)/(2) - (25)/(2) = 0](https://img.qammunity.org/2022/formulas/mathematics/high-school/6ofiqfvzejtfqkzw1kambe89o0rfv1czpv.png)
Both results make the equation true, so
![x = 0, -(5)/(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/5g1eqa1p8008y5w0eey8ruq9sjzm9uvnr2.png)