Question

15t2 + 7t = 2 solve

Answer 1

\[15t^{2}+7t=2\]

\[15t^{2}+7t-2=0\]
\[(3x+2)(5x -1)\]
\[x=\{\frac{-2}{3}, \frac{1}{5}\}\]

Answer 2

okay i see how you did it i guess i am just trying to make some sense of it

Answer 3

Any equation to the form of:
\[\Large a{x^2} + bx + c=0\]
Is a quadratic and can be solved by either factorising or using the quadratic formula. Note here that although it's laid out differently, you can just move the 2 to the other side (as jlvmrc has done):
\[\Large 15{t^2}{\rm{ }} + {\rm{ }}7t - 2 = 0\]
Then just plug the corresponding numbers into the quadratic formula and solve:
\[\Large \begin{array}{l}
\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
\frac{{ - 7 \pm \sqrt {{7^2} - 4(15 \times - 2)} }}{{2 \times 15}}\\
= \frac{{ - 7 \pm \sqrt {49 + 120} }}{{30}}\\
= \frac{{ - 7 \pm 13}}{{30}}\\
= \left\{ {{\rm{0}}{\rm{.2}}, - \frac{2}{3}} \right\}
\end{array}\]
And you end up with the roots. For a general quadratic with no specific attachment, both roots are generally the right answer, but if you're dealing with something tangible or conceptual - for instance, a physical quantity - then you need to take into account which is 'real' and which isn't (e.g. you can't have negative of a physical quantity, so if there's a negative and a positive root, you only take the positive).