Use the quadratic formula to solve for the roots in the following equation.4x2+5x + 2 = 2x2+7x-1

Answer:The roots of [tex]4 x^{2}+5 x+2=2 x^{2}+7 x-1[/tex]  are [tex]\frac{2+2 i \sqrt{5}}{2}[/tex] and [tex]\frac{2-2 i \sqrt{5}}{2}[/tex]Solution: The equation given is, [tex]4 x^{2}+5 x+2=2 x^{2}+7 x-1[/tex]Simplifying we get, [tex]4 x^{2}+5 x+2=2 x^{2}+7 x-1[/tex][tex]4x^2- 2x^2 +5x- 7x +2+1 = 0[/tex][tex](4x^2-2x^2) +(5x -7x) +(2+1) = 0[/tex][tex]2x^2 -2x +3 =0[/tex]We know that the quadratic formula to solve this, X has two values which are [tex]=\frac{(-b+\sqrt{b^{2}-4 a c})}{2 a} \text { and } \frac{-b-\sqrt{b^{2}-4 a c}}{2 a}[/tex]Here a= 2; b = -2; c = 3 So substituting the values we get, [tex]x=\frac{-(-2)+\sqrt{(-2)^{2}-4 \times 2 \times 3}}{2 \times 2}[/tex][tex]\Rightarrow\frac{2+\sqrt{4-24}}{4}=\frac{2+\sqrt{-20}}{4}=\frac{2+\sqrt{(-1) \times 4 \times 5}}{4}[/tex][tex]\Rightarrow=\frac{2+2 i \sqrt{5}}{2}[/tex]   [tex](\text { assuming }-\sqrt{-1}=\mathrm{i})[/tex]Again [tex]x=\frac{2-2 i \sqrt{5}}{2}[/tex]So, the roots are [tex]\frac{2+2 i \sqrt{5}}{2}[/tex] and [tex]\frac{2-2 i \sqrt{5}}{2}[/tex]