soccer ball profit\n$y = -6x^{2}+100x - 180$\nsuppose the store wants to earn a daily profit of $150 from…

Answer:

We need to solve the quadratic equation $-6x^{2}+100x - 180=150$ for $x$. First, rewrite it in standard form $ax^{2}+bx + c = 0$. Then use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ to find the values of $x$ (where $x$ represents the price of each soccer - ball).

Explanation:

Step1: Rewrite the equation in standard form

Set $y = 150$ in $y=-6x^{2}+100x - 180$. We get $-6x^{2}+100x-180 - 150=0$, which simplifies to $-6x^{2}+100x - 330 = 0$. Divide through by $-2$ to get $3x^{2}-50x + 165 = 0$. Here, $a = 3$, $b=-50$, and $c = 165$.

Step2: Apply the quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 3$, $b=-50$, and $c = 165$ into it. First, calculate the discriminant $\Delta=b^{2}-4ac=(-50)^{2}-4\times3\times165=2500 - 1980 = 520$. Then $x=\frac{50\pm\sqrt{520}}{6}=\frac{50\pm2\sqrt{130}}{6}=\frac{25\pm\sqrt{130}}{3}$. So the two possible values of $x$ (prices of soccer - balls) are $x=\frac{25+\sqrt{130}}{3}\approx\frac{25 + 11.4}{3}\approx12.13$ and $x=\frac{25-\sqrt{130}}{3}\approx\frac{25 - 11.4}{3}\approx4.53$. We may need to consider the context to determine which value (if any) is a reasonable price for a soccer - ball.