Determine the number of solutions for each quadratic equation.

(a) $5x^2-7x-8=0$

(b) $7x^2-10x+5=0$

(c) $25x^2-90x+81=0$

(d)$15x^2-8x+4=0$

The given equation is$5x^2-7x-8=0$.

Comparing the given equation with the standard form we have $a=5,b=-7,c=-8$.

$\begin{array}{rcl}{b}^2-4ac& =& {\left(-7\right)}^2-4\times 5\times \left(-8\right)\\ & =& 49+160\\ & =& 209>0\end{array}$

The discriminant is positive.

So, there are two real solutions.

The given equation is $7x^2-10x+5=0$.

Comparing the given equation with the standard form we have $a=7,b=-10,c=5$.

$\begin{array}{rcl}{b}^2-4ac& =& {\left(-10\right)}^2-4\times 7\times 5\\ & =& 100-140\\ & =& -40<0\end{array}$

The discriminant is negative.

So, there are no real solutions.

Hence, two complex solutions exist.

The given equation is$25x^2-90x+81=0$.

Comparing the given equation with the standard form we have $a=25,b=-90,c=81$.

$\begin{array}{rcl}{b}^2-4ac& =& {\left(-90\right)}^2-4\times 25\times 81\\ & =& 8100-8100\\ & =& 0\end{array}$

The discriminant is zero.

So, there are one real solution.

The given equation is$15x^2-8x+4=0$.

Comparing the given equation with the standard form we have $a=15,b=-8,c=4$.

$\begin{array}{rcl}{b}^2-4ac& =& {\left(-8\right)}^2-4\times 15\times 4\\ & =& 64-240\\ & =& -176<0\end{array}$

The discriminant is negative.

So, there is no real solution.

Hence, two complex solutions exist.