Answer:
[tex]\frac{2\sqrt3}{3} = x[/tex]
Step-by-step explanation:
So you want to use PEMDAS to do the equation in order. This means that you'll start by adding 6 and 1 and then multiplying that value by 2
[tex]2(7)-|-10| = 3x^2[/tex]
[tex]14-|-10|=3x^2[/tex]
The absolute value can be defined as the "distance" from 0 which is going to be positive. You can think of it as the positive value of a number so if you have |-b| it will become b, and even if it's already positive, it will remain positive. so the absolute value of -10 is 10 but then it becomes negative again because of the subtraction
[tex]14-10=3x^2[/tex]
[tex]4=3x^2[/tex]
Now you want to solve for x by dividing both sides by 3 and taking the square root of both sides to isolate x
[tex]\frac{4}{3} = x^2[/tex]
[tex]\sqrt{\frac{4}{3}}=x[/tex]
You can distribute the square root across division since [tex](\frac{a}{b})^c = \frac{a^c}{b^c}[/tex] and the square root can be defined as [tex]x^{0.5}[/tex] since [tex]a^b * a^c = a^{b+c}[/tex] because if you spread it out it's really just [tex](a * a * a ... b\ amount \ of \ times) * (a * a * a ... c\ amount\ of\ times)[/tex] which can be simplified to (a * a * a... (b + c) amount of times) if that makes any sense. Anyways using this property you'll get [tex]a^{0.50} * a^{0.50} = a^{0.50 + 0.50} = a^1 = a[/tex]. If you look at it you'll notice a^0.50 multiplied by it self gives you a... sounds like the square root, which it is. So now the equation becomes
[tex]\frac{\sqrt{4}}{\sqrt{3}}=x[/tex]
[tex]\frac{2}{\sqrt3}[/tex]
Now if you look at the equation you'll notice there's a radical in the denominator, which can be rationalized by multiplying by sqrt(3)\sqrt(3) which is 1 except it makes the denominator 3 and keeps the original value
[tex]\frac{2}{\sqrt3} * \frac{\sqrt3}{\sqrt3}=x\\\frac{2\sqrt{3}}{3} = x[/tex]
