Solve (-2x+4)^2*-4(5)(-2)>0

$$( - 2 x + 4 ) ^ { 2 } \cdot - 4 ( 5 ) ( - 2 ) > 0$$

Solve for x

$$x\neq 2$$

Steps Using the Quadratic Formula

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

$$\left(4x^{2}-16x+16\right)\left(-4\right)\times 5\left(-2\right)>0$$

Multiply -4 and 5 to get -20.

$$\left(4x^{2}-16x+16\right)\left(-20\right)\left(-2\right)>0$$

Multiply -20 and -2 to get 40.

$$\left(4x^{2}-16x+16\right)\times 40>0$$

Use the distributive property to multiply 4x^{2}-16x+16 by 40.

$$160x^{2}-640x+640>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$160x^{2}-640x+640=0$$

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 160 for a, -640 for b, and 640 for c in the quadratic formula.

$$x=\frac{-\left(-640\right)±\sqrt{\left(-640\right)^{2}-4\times 160\times 640}}{2\times 160}$$

Do the calculations.

$$x=\frac{640±0}{320}$$

Solutions are the same.

$$x=2$$

Rewrite the inequality by using the obtained solutions.

$$160\left(x-2\right)^{2}>0$$

Inequality holds for x\neq 2.

$$x\neq 2$$