We are asked to justify the method and solve the given quadratic equations.
Part A:
[tex]2x(x+1.5)=-1[/tex]
Let us solve the above quadratic equation using the quadratic formula
First, expand the equation to convert it into the standard form.
[tex]\begin{gathered} 2x(x+1.5)=-1 \\ 2x^2+3x=-1 \\ 2x^2+3x+1=0 \end{gathered}[/tex]
So, the coefficients of the quadratic equation are
a = 2
b = 3
c = 1
The quadratic formula is given by
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{-3\pm\sqrt[]{3^2-4(2)(1)}}{2(2)} \\ x=\frac{-3\pm\sqrt[]{9^{}-8}}{4} \\ x=\frac{-3\pm\sqrt[]{1}}{4} \\ x=\frac{-3\pm1}{4} \\ x=\frac{-3+1}{4},\; \frac{-3-1}{4}\; \\ x=\frac{-2}{4},\; \frac{-4}{4}\; \\ x=-0.5,\; -1\; \end{gathered}[/tex]
Therefore, the solution is x = -0.5 and x = -1
