Answer :
To apply the quadratic formula, the right side must be equal to 0.
[tex]2w^2-28w=-98 \\ 2w^2-28w+98=0[/tex]
Now you can apply the quadratic formula [tex]\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex] to find the solution for w.
[tex]\frac{-(-28)\±\sqrt{(-28)^2-4(2)(98)}}{2(2)} = \frac{28\±\sqrt{784-784}}{4} = \frac{28}4 = \boxed{7} [/tex]
(The quadratic formula is highly unnecessary in this problem, however. You could easily divide all values by 2 to get w²-14w+49=0, which factors into (w-7)²=0, for which 7 is the only solution which puts a factor of the quadratic equal to 0.)
[tex]2w^2-28w=-98 \\ 2w^2-28w+98=0[/tex]
Now you can apply the quadratic formula [tex]\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex] to find the solution for w.
[tex]\frac{-(-28)\±\sqrt{(-28)^2-4(2)(98)}}{2(2)} = \frac{28\±\sqrt{784-784}}{4} = \frac{28}4 = \boxed{7} [/tex]
(The quadratic formula is highly unnecessary in this problem, however. You could easily divide all values by 2 to get w²-14w+49=0, which factors into (w-7)²=0, for which 7 is the only solution which puts a factor of the quadratic equal to 0.)
