Solve 17c^2+(3+2sqrt{3})c+36*4=1

$$17 c ^ { 2 } + ( 3 + 2 \sqrt { 3 } ) c + 36 \times 4 = 1$$

Solve for c

$$c=\frac{i\sqrt{9703-12\sqrt{3}}}{34}-\frac{\sqrt{3}}{17}-\frac{3}{34}\approx -0.190120636+2.894066145i$$

$$c=-\frac{i\sqrt{9703-12\sqrt{3}}}{34}-\frac{\sqrt{3}}{17}-\frac{3}{34}\approx -0.190120636-2.894066145i$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$17c^{2}+\left(2\sqrt{3}+3\right)c+144=1$$

Subtract 1 from both sides of the equation.

$$17c^{2}+\left(2\sqrt{3}+3\right)c+144-1=1-1$$

Subtracting 1 from itself leaves 0.

$$17c^{2}+\left(2\sqrt{3}+3\right)c+144-1=0$$

Subtract 1 from 144.

$$17c^{2}+\left(2\sqrt{3}+3\right)c+143=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 17 for a, 3+2\sqrt{3} for b, and 143 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(2\sqrt{3}+3\right)±\sqrt{\left(2\sqrt{3}+3\right)^{2}-4\times 17\times 143}}{2\times 17}$$

Square 3+2\sqrt{3}.

$$c=\frac{-\left(2\sqrt{3}+3\right)±\sqrt{12\sqrt{3}+21-4\times 17\times 143}}{2\times 17}$$

Multiply -4 times 17.

$$c=\frac{-\left(2\sqrt{3}+3\right)±\sqrt{12\sqrt{3}+21-68\times 143}}{2\times 17}$$

Multiply -68 times 143.

$$c=\frac{-\left(2\sqrt{3}+3\right)±\sqrt{12\sqrt{3}+21-9724}}{2\times 17}$$

Add 21+12\sqrt{3} to -9724.

$$c=\frac{-\left(2\sqrt{3}+3\right)±\sqrt{12\sqrt{3}-9703}}{2\times 17}$$

Take the square root of -9703+12\sqrt{3}.

$$c=\frac{-\left(2\sqrt{3}+3\right)±i\sqrt{9703-12\sqrt{3}}}{2\times 17}$$

Multiply 2 times 17.

$$c=\frac{-2\sqrt{3}-3±i\sqrt{9703-12\sqrt{3}}}{34}$$

Now solve the equation c=\frac{-2\sqrt{3}-3±i\sqrt{9703-12\sqrt{3}}}{34} when ± is plus. Add -3-2\sqrt{3} to i\sqrt{9703-12\sqrt{3}}.

$$c=\frac{-2\sqrt{3}-3+i\sqrt{9703-12\sqrt{3}}}{34}$$

Divide -3-2\sqrt{3}+i\sqrt{9703-12\sqrt{3}} by 34.

$$c=\frac{i\sqrt{9703-12\sqrt{3}}}{34}-\frac{\sqrt{3}}{17}-\frac{3}{34}$$

Now solve the equation c=\frac{-2\sqrt{3}-3±i\sqrt{9703-12\sqrt{3}}}{34} when ± is minus. Subtract i\sqrt{9703-12\sqrt{3}} from -3-2\sqrt{3}.

$$c=\frac{-i\sqrt{9703-12\sqrt{3}}-2\sqrt{3}-3}{34}$$

Divide -3-2\sqrt{3}-i\sqrt{9703-12\sqrt{3}} by 34.

$$c=-\frac{i\sqrt{9703-12\sqrt{3}}}{34}-\frac{\sqrt{3}}{17}-\frac{3}{34}$$

The equation is now solved.

$$c=\frac{i\sqrt{9703-12\sqrt{3}}}{34}-\frac{\sqrt{3}}{17}-\frac{3}{34} c=-\frac{i\sqrt{9703-12\sqrt{3}}}{34}-\frac{\sqrt{3}}{17}-\frac{3}{34}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$17c^{2}+\left(2\sqrt{3}+3\right)c+144=1$$

Subtract 144 from both sides of the equation.

$$17c^{2}+\left(2\sqrt{3}+3\right)c+144-144=1-144$$

Subtracting 144 from itself leaves 0.

$$17c^{2}+\left(2\sqrt{3}+3\right)c=1-144$$

Subtract 144 from 1.

$$17c^{2}+\left(2\sqrt{3}+3\right)c=-143$$

Divide both sides by 17.

$$\frac{17c^{2}+\left(2\sqrt{3}+3\right)c}{17}=-\frac{143}{17}$$

Dividing by 17 undoes the multiplication by 17.

$$c^{2}+\frac{2\sqrt{3}+3}{17}c=-\frac{143}{17}$$

Divide \frac{3+2\sqrt{3}}{17}, the coefficient of the x term, by 2 to get \frac{3}{34}+\frac{\sqrt{3}}{17}. Then add the square of \frac{3}{34}+\frac{\sqrt{3}}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}+\frac{2\sqrt{3}+3}{17}c+\left(\frac{\sqrt{3}}{17}+\frac{3}{34}\right)^{2}=-\frac{143}{17}+\left(\frac{\sqrt{3}}{17}+\frac{3}{34}\right)^{2}$$

Square \frac{3}{34}+\frac{\sqrt{3}}{17}.

$$c^{2}+\frac{2\sqrt{3}+3}{17}c+\frac{3\sqrt{3}}{289}+\frac{21}{1156}=-\frac{143}{17}+\frac{3\sqrt{3}}{289}+\frac{21}{1156}$$

Add -\frac{143}{17} to \frac{21}{1156}+\frac{3\sqrt{3}}{289}.

$$c^{2}+\frac{2\sqrt{3}+3}{17}c+\frac{3\sqrt{3}}{289}+\frac{21}{1156}=\frac{3\sqrt{3}}{289}-\frac{9703}{1156}$$

Factor c^{2}+\frac{2\sqrt{3}+3}{17}c+\frac{3\sqrt{3}}{289}+\frac{21}{1156}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c+\frac{\sqrt{3}}{17}+\frac{3}{34}\right)^{2}=\frac{3\sqrt{3}}{289}-\frac{9703}{1156}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c+\frac{\sqrt{3}}{17}+\frac{3}{34}\right)^{2}}=\sqrt{\frac{3\sqrt{3}}{289}-\frac{9703}{1156}}$$

Simplify.

$$c+\frac{\sqrt{3}}{17}+\frac{3}{34}=\frac{i\sqrt{9703-12\sqrt{3}}}{34} c+\frac{\sqrt{3}}{17}+\frac{3}{34}=-\frac{i\sqrt{9703-12\sqrt{3}}}{34}$$

Subtract \frac{3}{34}+\frac{\sqrt{3}}{17} from both sides of the equation.

$$c=\frac{i\sqrt{9703-12\sqrt{3}}}{34}-\frac{\sqrt{3}}{17}-\frac{3}{34} c=-\frac{i\sqrt{9703-12\sqrt{3}}}{34}-\frac{\sqrt{3}}{17}-\frac{3}{34}$$