Answer:
This equation does have real solutions contrary to your question. They are [tex]\frac{-9 + \sqrt{101}} {-10}[/tex] OR [tex]\frac{-9 - \sqrt{101}} {-10}[/tex].
Step-by-step explanation:
In order to prove this has real number solutions, we simply need to use the quadratic formula. To do this, we first need to solve for 0.
2x^2 + 18x = 10
2x^2 + 18x - 10 = 0
Now that we have this and they are all factors of 2, we can divide everything by 2 for ease.
2x^2 + 18x - 10 = 0
x^2 + 9x - 5 = 0
Now we can use the quadratic equation to solve. Remember that A = 1 due to the coefficient of x^2, B = 9 as the coefficient of x and C = -5 as the constant.
[tex]\frac{-b +/- \sqrt{b^{2} - 4ac}} {2a}[/tex]
[tex]\frac{-9 +/- \sqrt{9^{2} - 4(1)(-5)}} {2(-5)}[/tex]
[tex]\frac{-9 +/- \sqrt{81 + 20}} {-10}[/tex]
[tex]\frac{-9 +/- \sqrt{101}} {-10}[/tex]
[tex]\frac{-9 + \sqrt{101}} {-10}[/tex] OR [tex]\frac{-9 - \sqrt{101}} {-10}[/tex]
