Solve (2x+4)^2-5x(7-3x)(7+3x)-(3x-2)^2-40x^2=-205

$$( 2 x + 4 ) ^ { 2 } - 5 x ( 7 - 3 x ) ( 7 + 3 x ) - ( 3 x - 2 ) ^ { 2 } - 40 x ^ { 2 } = - 205$$

Solve for x

$$x = \frac{\sqrt{1085}}{15} \approx 2.195955879$$

$$x = -\frac{\sqrt{1085}}{15} \approx -2.195955879$$

$$x=1$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+4\right)^{2}.

$$4x^{2}+16x+16-5x\left(7-3x\right)\left(7+3x\right)-\left(3x-2\right)^{2}-40x^{2}=-205$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$4x^{2}+16x+16-5x\left(7-3x\right)\left(7+3x\right)-\left(9x^{2}-12x+4\right)-40x^{2}=-205$$

To find the opposite of 9x^{2}-12x+4, find the opposite of each term.

$$4x^{2}+16x+16-5x\left(7-3x\right)\left(7+3x\right)-9x^{2}+12x-4-40x^{2}=-205$$

Combine -9x^{2} and -40x^{2} to get -49x^{2}.

$$4x^{2}+16x+16-5x\left(7-3x\right)\left(7+3x\right)-49x^{2}+12x-4=-205$$

Add 205 to both sides.

$$4x^{2}+16x+16-5x\left(7-3x\right)\left(7+3x\right)-49x^{2}+12x-4+205=0$$

Add -4 and 205 to get 201.

$$4x^{2}+16x+16-5x\left(7-3x\right)\left(7+3x\right)-49x^{2}+12x+201=0$$

Use the distributive property to multiply -5x by 7-3x.

$$4x^{2}+16x+16+\left(-35x+15x^{2}\right)\left(7+3x\right)-49x^{2}+12x+201=0$$

Use the distributive property to multiply -35x+15x^{2} by 7+3x and combine like terms.

$$4x^{2}+16x+16-245x+45x^{3}-49x^{2}+12x+201=0$$

Combine 16x and -245x to get -229x.

$$4x^{2}-229x+16+45x^{3}-49x^{2}+12x+201=0$$

Combine 4x^{2} and -49x^{2} to get -45x^{2}.

$$-45x^{2}-229x+16+45x^{3}+12x+201=0$$

Combine -229x and 12x to get -217x.

$$-45x^{2}-217x+16+45x^{3}+201=0$$

Add 16 and 201 to get 217.

$$-45x^{2}-217x+217+45x^{3}=0$$

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

$$45x^{3}-45x^{2}-217x+217=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 217 and q divides the leading coefficient 45. List all candidates \frac{p}{q}.

$$±\frac{217}{45},±\frac{217}{15},±\frac{217}{9},±\frac{217}{5},±\frac{217}{3},±217,±\frac{31}{45},±\frac{31}{15},±\frac{31}{9},±\frac{31}{5},±\frac{31}{3},±31,±\frac{7}{45},±\frac{7}{15},±\frac{7}{9},±\frac{7}{5},±\frac{7}{3},±7,±\frac{1}{45},±\frac{1}{15},±\frac{1}{9},±\frac{1}{5},±\frac{1}{3},±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 45x^{3}-45x^{2}-217x+217 by x-1 to get 45x^{2}-217. Solve the equation where the result equals to 0.

$$45x^{2}-217=0$$

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 45 for a, 0 for b, and -217 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 45\left(-217\right)}}{2\times 45}$$

Do the calculations.

$$x=\frac{0±6\sqrt{1085}}{90}$$

Solve the equation 45x^{2}-217=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{1085}}{15} x=\frac{\sqrt{1085}}{15}$$

List all found solutions.

$$x=1 x=-\frac{\sqrt{1085}}{15} x=\frac{\sqrt{1085}}{15}$$