Answer:
The roots are 7 and -6.
Keys:
- [tex]ax^2+bx+c=0[/tex]
- [tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
- [tex]-(-a)^(^n^) = a^(^n^)[/tex]
- [tex]\sqrt[n]{a}^n = a[/tex]
- [tex]1^a = 1[/tex]
When you see a ± in a quadratic equation, you will have multiple solutions; more than one.
Step-by-step explanation:
solve the main expression
[tex]x^2-x-42=0\\x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \left(-42\right)}}{2\cdot \:1}\\\rightarrow \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \left(-42\right)}\\=\sqrt{\left(-1\right)^2+4\cdot \:1\cdot \:42}\\\rightarrow \left(-1\right)^2\\=1^2\\=1\\\rightarrow 4\cdot \:1\cdot \:42\\=168\\=\sqrt{1+168}\\=\sqrt{169}\\=\sqrt{13^2}\\=13\\x_{1,\:2}=\frac{-\left(-1\right)\pm \:13}{2\cdot \:1}[/tex]
solve for x₁
[tex]\frac{-\left(-1\right)+13}{2\cdot \:1}\\=\frac{1+13}{2\cdot \:1}\\=\frac{14}{2\cdot \:1}\\=\frac{14}{2}\\=7[/tex]
solve for x₂
[tex]\frac{-\left(-1\right)-13}{2\cdot \:1}\\=\frac{1-13}{2\cdot \:1}\\=\frac{-12}{2\cdot \:1}\\=\frac{-12}{2}\\=-\frac{12}{2}\\=-6[/tex]
