Answer:
[tex]x=\frac{3}{4},\:x=-1[/tex]
Keys:
For this problem, you need the quadratic formula(listed below).
When you see ± in a quadratic equation, you must know there is going to be at least 2 solutions.
Step-by-step explanation:
solving for x₁ and x₂
[tex]4x^2+x=3\\4x^2+x-3=3-3\\4x^2+x-3=0\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot 4\left(-3\right)}}{2\cdot 4}\\[/tex]
[tex]1^2=1\\=\sqrt{1-4\cdot \:4\left(-3\right)}\\=\sqrt{1+4\cdot \:4\cdot \:3}\\=\sqrt{1+48}\\=\sqrt{49}\\=\sqrt{7^2}\\\sqrt{7^2}=7\\=7[/tex]
[tex]x_{1,\:2}=\frac{-1\pm \:7}{2\cdot \:4}\\x_1=\frac{-1+7}{2\cdot \:4},\:x_2=\frac{-1-7}{2\cdot \:4}\\[/tex]
solve for x₁
[tex]\frac{-1+7}{2\cdot \:4}[/tex]
[tex]=\frac{6}{2\cdot \:4}[/tex]
[tex]=\frac{6}{8}[/tex]
[tex]= \frac{6\div2}{8\div2}[/tex]
[tex]=\frac{3}{4}[/tex]
solve for x₂
[tex]\frac{-1-7}{2\cdot \:4}[/tex]
[tex]=\frac{-8}{2\cdot \:4}[/tex]
[tex]=\frac{-8}{8}[/tex]
[tex]=-\frac{8}{8}[/tex]
[tex]=-1[/tex]
Hope this helps!
