Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{t}^{2}ha{y}^{3}-2{y}^{2}-7y+4=0\]
Add \(2{y}^{2}\) to both sides.
\[{t}^{2}ha{y}^{3}-7y+4=2{y}^{2}\]
Add \(7y\) to both sides.
\[{t}^{2}ha{y}^{3}+4=2{y}^{2}+7y\]
Subtract \(4\) from both sides.
\[{t}^{2}ha{y}^{3}=2{y}^{2}+7y-4\]
Divide both sides by \({t}^{2}\).
\[ha{y}^{3}=\frac{2{y}^{2}+7y-4}{{t}^{2}}\]
Split the second term in \(2{y}^{2}+7y-4\) into two terms.
Multiply the coefficient of the first term by the constant term.
\[2\times -4=-8\]
Ask: Which two numbers add up to \(7\) and multiply to \(-8\)?
Split \(7y\) as the sum of \(8y\) and \(-y\).
\[2{y}^{2}+8y-y-4\]
\[ha{y}^{3}=\frac{2{y}^{2}+8y-y-4}{{t}^{2}}\]
Factor out common terms in the first two terms, then in the last two terms.
\[ha{y}^{3}=\frac{2y(y+4)-(y+4)}{{t}^{2}}\]
Factor out the common term \(y+4\).
\[ha{y}^{3}=\frac{(y+4)(2y-1)}{{t}^{2}}\]
Divide both sides by \(h\).
\[a{y}^{3}=\frac{\frac{(y+4)(2y-1)}{{t}^{2}}}{h}\]
Simplify \(\frac{\frac{(y+4)(2y-1)}{{t}^{2}}}{h}\) to \(\frac{(y+4)(2y-1)}{{t}^{2}h}\).
\[a{y}^{3}=\frac{(y+4)(2y-1)}{{t}^{2}h}\]
Divide both sides by \({y}^{3}\).
\[a=\frac{\frac{(y+4)(2y-1)}{{t}^{2}h}}{{y}^{3}}\]
Simplify \(\frac{\frac{(y+4)(2y-1)}{{t}^{2}h}}{{y}^{3}}\) to \(\frac{(y+4)(2y-1)}{{t}^{2}h{y}^{3}}\).
\[a=\frac{(y+4)(2y-1)}{{t}^{2}h{y}^{3}}\]
a=((y+4)*(2*y-1))/(t^2*h*y^3)
