Solve 5x^2-3x+(2x+3)^2>2x(x+7)

$$5 x ^ { 2 } - 3 x + ( 2 x + 3 ) ^ { 2 } > 2 x ( x + 7 )$$

Solve for x

$$x\in \mathrm{R}$$

Steps Using the Quadratic Formula

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

$$5x^{2}-3x+4x^{2}+12x+9>2x\left(x+7\right)$$

Combine 5x^{2} and 4x^{2} to get 9x^{2}.

$$9x^{2}-3x+12x+9>2x\left(x+7\right)$$

Combine -3x and 12x to get 9x.

$$9x^{2}+9x+9>2x\left(x+7\right)$$

Use the distributive property to multiply 2x by x+7.

$$9x^{2}+9x+9>2x^{2}+14x$$

Subtract 2x^{2} from both sides.

$$9x^{2}+9x+9-2x^{2}>14x$$

Combine 9x^{2} and -2x^{2} to get 7x^{2}.

$$7x^{2}+9x+9>14x$$

Subtract 14x from both sides.

$$7x^{2}+9x+9-14x>0$$

Combine 9x and -14x to get -5x.

$$7x^{2}-5x+9>0$$

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.

$$7x^{2}-5x+9=0$$

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 7 for a, -5 for b, and 9 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 7\times 9}}{2\times 7}$$

Do the calculations.

$$x=\frac{5±\sqrt{-227}}{14}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 7x^{2}-5x+9 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$7\times 0^{2}-5\times 0+9=9$$

The value of the expression 7x^{2}-5x+9 is always positive. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$