[tex]30z^{2} =18-7z[/tex]

Solve With Quadratic Formula
Eg: For The Quadratic Equation Of The Form [tex]\bold{ax^2+bx+c=0}[/tex] The Solutions Are [tex]\bold{x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}[/tex]

For [tex]\bold{a=30,\:b=7,\:c=-18:\quad z_{1,\:2}=\frac{-7\pm \sqrt{7^2-4\cdot \:30\left(-18\right)}}{2\cdot \:30}}[/tex]

[tex]\bold{\frac{-7+\sqrt{7^2-4\cdot \:30\left(-18\right)}}{2\cdot \:30}: \ \frac{2}{3} }[/tex]

[tex]\bold{\frac{-7-\sqrt{7^2-4\cdot \:30\left(-18\right)}}{2\cdot \:30}: \ -\frac{9}{10} }[/tex]

Solutions:

[tex]\bold{z=\frac{2}{3},\:z=-\frac{9}{10}}[/tex]

- M

kudzordzifrancis kudzordzifrancis · Answer 2 · 2019-04-23T18:10:44+02:00

Answer:

[tex]z=-\frac{9}{10}[/tex] or [tex]z=\frac{2}{3}[/tex]

Step-by-step explanation:

We want to solve;

[tex]30z^2=18-7z[/tex]

We rewrite in the form;

[tex]az^2+bz+c=0[/tex]

[tex]30z^2+7z-18=0[/tex]

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

The solution is given by;

[tex]z=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]

Substitute to obtain;

[tex]z=\frac{-7\pm \sqrt{7^2-4(30)(-18)} }{2(30)}[/tex]

[tex]z=\frac{-7\pm \sqrt{2209} }{2(30)}[/tex]

[tex]z=\frac{-7\pm 47 }{60}[/tex]

[tex]z=-\frac{9}{10}[/tex] or [tex]z=\frac{2}{3}[/tex]