Determine the number of solutions to each quadratic equation:

(a) $4x^2-5x+16=0$

(b) $36y^2+36y+9=0$

(c) $6m^2+3m-5=0$

(d)$18n^2-7n+3=0$

We have to find the number of solutions for the given equations.

Given, $4x^2-5x+16=0$

Here, $a=4,b=-5,c=16$.

Finding the discriminant.

$$\begin{gathered} D=b^2-4ac \\ ={(-5)}^2-4\left(4\right)\left(16\right) \\ =25-256 \\ =-231 \end{gathered}$$

Since the discriminant is less than zero there are no real solutions to the equation.

Given, $36y^2+36y+9=0$

Here, $a=36,b=36,c=9$

Finding the discriminant.

$$\begin{gathered} D=b^2-4ac \\ ={\left(36\right)}^2-4\left(36\right)\left(9\right) \\ =1296-1296 \\ =0 \end{gathered}$$

Since the discriminant is zero, there is one solution to the equation.

Given, $6m^2+3m-5=0$

Here, $a=6,b=3,c=-5$

Finding the discriminant.

$$\begin{gathered} D=b^2-4ac \\ ={\left(3\right)}^2-4\left(6\right)(-5) \\ =9+120 \\ =129 \end{gathered}$$

Since the discriminant is positive there are two solutions to the equation.

Given, $18n^2-7n+3=0$.

Here $a=18,b=-7,c=3$

Finding the discriminant.

$$\begin{gathered} D=b^2-4ac \\ ={(-7)}^2-2\left(18\right)\left(3\right) \\ =49+108 \\ =157 \end{gathered}$$

Since the discriminant is positive there are two solutions to the equation.