Solve 6y^3=4-13y^2

$$6 y ^ { 3 } = 4 - 13 y ^ { 2 }$$

Solve for y

$$y=\frac{1}{2}=0.5$$

$$y=-\frac{2}{3}\approx -0.666666667$$

$$y=-2$$

Solution Steps

Subtract 4 from both sides.

$$6y^{3}-4=-13y^{2}$$

Add 13y^{2} to both sides.

$$6y^{3}-4+13y^{2}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$6y^{3}+13y^{2}-4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.

$$±\frac{2}{3},±\frac{4}{3},±2,±4,±\frac{1}{3},±1,±\frac{1}{6},±\frac{1}{2}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$y=-2$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 6y^{3}+13y^{2}-4 by y+2 to get 6y^{2}+y-2. Solve the equation where the result equals to 0.

$$6y^{2}+y-2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 6 for a, 1 for b, and -2 for c in the quadratic formula.

$$y=\frac{-1±\sqrt{1^{2}-4\times 6\left(-2\right)}}{2\times 6}$$

Do the calculations.

$$y=\frac{-1±7}{12}$$

Solve the equation 6y^{2}+y-2=0 when ± is plus and when ± is minus.

$$y=-\frac{2}{3} y=\frac{1}{2}$$

List all found solutions.

$$y=-2 y=-\frac{2}{3} y=\frac{1}{2}$$