Question

`30z^(2) =18-7z`

Answer 1

`\boxed{\bold{z=(2)/(3),\:z=-(9)/(10)}}`

Step By Step Explanation:

Add 7z To Both Sides

`\bold{30z^2+7z=18-7z+7z}`

Simplify

`\bold{30z^2+7z=18}`

Subtract 18 From Both Sides

`\bold{30z^2+7z-18=18-18}`

Simplify

`\bold{30z^2+7z-18=0}`

• Solve With Quadratic Formula
• Eg: For The Quadratic Equation Of The Form
  `\bold{ax^2+bx+c=0}` The Solutions Are
  `\bold{x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)}`

• For
  `\bold{a=30,\:b=7,\:c=-18:\quad z_(1,\:2)=(-7\pm √(7^2-4\cdot \:30\left(-18\right)))/(2\cdot \:30)}`

`\bold{(-7+√(7^2-4\cdot \:30\left(-18\right)))/(2\cdot \:30): \ (2)/(3) }`

`\bold{(-7-√(7^2-4\cdot \:30\left(-18\right)))/(2\cdot \:30): \ -(9)/(10) }`

Solutions:

`\bold{z=(2)/(3),\:z=-(9)/(10)}`

- M

Answer 2

`z=-(9)/(10)` or
`z=(2)/(3)`

Explanation:

We want to solve;

`30z^2=18-7z`

We rewrite in the form;

`az^2+bz+c=0`

`30z^2+7z-18=0`

where a=30,b=7 and c=-18

The solution is given by;

`z=(-b\pm √(b^2-4ac) )/(2a)`

Substitute to obtain;

`z=(-7\pm √(7^2-4(30)(-18)) )/(2(30))`

`z=(-7\pm √(2209) )/(2(30))`

`z=(-7\pm 47 )/(60)`

`z=-(9)/(10)` or
`z=(2)/(3)`