Answer:
[tex]-\frac{3}{8} +\frac{\sqrt[]{23}i }{8},-\frac{3}{8} -\frac{\sqrt[]{23}i }{8}[/tex]
or
[tex]-0.375+0.599479i,-0.375-0.599479i[/tex]
Step-by-step explanation:
[tex]4r^2+3r+10=8[/tex]
Bring the 8 to the left side so that we equal the equation to 0. To do this, simply substract 8 on both sides.
[tex]4r^2+3r+10-8=8-8\\4r^2+3r+2=0[/tex]
Where;
[tex]a=4\\b=3\\c=2[/tex]
[tex]Formula: x=\frac{-b\frac{+}{}\sqrt[]{b^2-4ac} }{2a}[/tex]
Replace. Let x be r
[tex]r=\frac{-3\frac{+}{}\sqrt[]{(3)^2-4(4)(2)} }{2(4)}[/tex]
[tex]r=\frac{-3\frac{+}{}\sqrt[]{9-32} }{8}[/tex]
[tex]r=\frac{-3\frac{+}{}\sqrt[]{-23} }{8}[/tex]
It has no real solution because the square root is negative.
We can say that,
[tex]r=-\frac{3}{8} \frac{+}{}\frac{\sqrt[]{23}*\sqrt[]{-1} }{8}[/tex]
[tex]r=-\frac{3}{8} \frac{+}{}\frac{\sqrt[]{23}i }{8}[/tex]
[tex]r_1=-0.375+0.599479i\\r_2=-0.375-0.599479i[/tex]