Answer
$3\text{ and }5$.
Work Step by Step
${{x}^{2}}\left( 3x+4 \right)=4x\left( x-1 \right)+15$.
Rewrite the expression ${{x}^{2}}\left( 3x+4 \right)=4x\left( x-1 \right)+15$:
$\begin{align}
& x\left( 3x+4 \right)=4x\left( x-1 \right)+15 \\
& 3{{x}^{2}}+4x=4{{x}^{2}}-4x+15 \\
\end{align}$
Now, subtract $4{{x}^{2}}$ on both sides:
$\begin{align}
& 3{{x}^{2}}+4x=4{{x}^{2}}-4x+15 \\
& 3{{x}^{2}}+4x-4{{x}^{2}}=4{{x}^{2}}-4{{x}^{2}}-4x+15\,\,\,
\end{align}$
Combine the like terms as,
$\begin{align}
& 3{{x}^{2}}+4x-4{{x}^{2}}=4{{x}^{2}}-4{{x}^{2}}-4x+15\,\,\, \\
& 3{{x}^{2}}+4x-4{{x}^{2}}=-4x+15\,\,\, \\
& 4x-{{x}^{2}}=-4x+15\,\,\,
\end{align}$
Now, add $4x$ on both sides as,
$\begin{align}
& 4x-{{x}^{2}}=-4x+15\,\, \\
& 4x-{{x}^{2}}+4x=4x-4x+15\,\, \\
& 8x-{{x}^{2}}=15 \\
\end{align}$
Now, subtract $15$ on both sides as,
$\begin{align}
& 8x-{{x}^{2}}=15 \\
& 8x-{{x}^{2}}-15=15-15 \\
& 8x-{{x}^{2}}-15=0 \\
& {{x}^{2}}-8x+15=0
\end{align}$
Compare the expression with the equation $a{{x}^{2}}+bx+c=0$.
Thus,
$a=1,b=-8,c=15$
Now, place these values in the quadratic equation formula,
$\begin{align}
& x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& =\frac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\cdot 1\cdot 15}}{2\cdot 1} \\
& =\frac{8\pm \sqrt{64-60}}{2} \\
& =\frac{8\pm \sqrt{4}}{2}
\end{align}$
Thus:
$\begin{align}
& x=\frac{8\pm \sqrt{4}}{2} \\
& x=\frac{8\pm 2}{2} \\
\end{align}$
Now, from the principle of squares property,
$x=\frac{8+2}{2}$ or $x=\frac{8-2}{2}$
Now, consider the value $x=\frac{8-2}{2}$:
$\begin{align}
& x=\frac{8-2}{2} \\
& =\frac{6}{2} \\
& =3
\end{align}$
Now consider the value $x=\frac{8+2}{2}$:
$\begin{align}
& x=\frac{8+2}{2} \\
& =\frac{10}{2} \\
& =5
\end{align}$
