Answer:
The roots are
[tex]x1= \frac{-3+\sqrt{233}}{4}[/tex]
[tex]x2= \frac{-3-\sqrt{233}}{4}[/tex]
Step-by-step explanation:
we have
[tex]2x^{2}+3x-15=13[/tex]
so
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]2x^{2}+3x=13+15[/tex]
[tex]2x^{2}+3x=28[/tex]
Factor the leading coefficient
[tex]2(x^{2}+(3/2)x)=28[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]2(x^{2}+(3/2)x+(9/16))=28+(9/8)[/tex]
[tex]2(x^{2}+(3/2)x+(9/16))=233/8[/tex]
[tex](x^{2}+(3/2)x+(9/16))=233/16[/tex]
Rewrite as perfect squares
[tex](x+(3/4))^{2}=233/16[/tex]
square root both sides
[tex](x+\frac{3}{4})=(+/-)\sqrt{\frac{233}{16}}\\ \\(x+\frac{3}{4})=(+/-)\frac{\sqrt{233}}{4}\\ \\x= -\frac{3}{4}(+/-)\frac{\sqrt{233}}{4}[/tex]
[tex]x1= -\frac{3}{4}(+)\frac{\sqrt{233}}{4}=\frac{-3+\sqrt{233}}{4}[/tex]
[tex]x2= -\frac{3}{4}(-)\frac{\sqrt{233}}{4}=\frac{-3-\sqrt{233}}{4}[/tex]
