How do you solve ${{x}^{2}}+3x-4=0$ graphically and algebraically? \[\]

Hint: We use the completing square method to solve the quadratic equation$a{{x}^{2}}+bx+c=0$. We first take $c$ to the right hand side of the equation then $a\ne 1$ we divide both sides by $a$. We then add ${{\left( \dfrac{-b}{2a} \right)}^{2}}$ both sides. We make a complete square and then take the square root on both sides. We use the information that $y=a{{x}^{2}}+bx+c\left( a>0 \right)$is the graph of an upward parabola with vertex at $x=\dfrac{-b}{2a}$ and solutions are at the point of intersection of the curve $y=a{{x}^{2}}+bx+c$ with $x-$axis. \[\]

Complete step-by-step solution:
We are given the quadratic equation ${{x}^{2}}+3x-4=0$ in the question; we compare it with general quadratic equation $a{{x}^{2}}+bx+c=0$ to find $a=1,b=3,c=4$. We follows steps of completing the square method and take $c$ to the right hand side to have
\[{{x}^{2}}+3x=4\]
Since $a=1$ we do not need to divide both sides .We add ${{\left( \dfrac{-b}{2a} \right)}^{2}}={{\left( \dfrac{\left( -3 \right)}{2\times 1} \right)}^{2}}={{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{9}{4}$ both side sides of the above equation to have
$\Rightarrow {{x}^{2}}-3x+\dfrac{9}{4}=4+\dfrac{9}{4}$
Now we shall make a complete square using the terms on the left-hand side of the above step. Let us have;
$\Rightarrow {{\left( x \right)}^{2}}-2\times x\times \dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{16+9}{4}$
We use the algebraic identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ in the left hand side of the above step to have the complete square as
\[\Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{25}{4}\]
 We take square root both sides of the above step to have;
\[\begin{align}
  & \Rightarrow \left( x-\dfrac{3}{2} \right)=\pm \dfrac{5}{2} \\
 & \Rightarrow x=\dfrac{3}{2}\pm \dfrac{5}{2} \\
 & \Rightarrow x=\dfrac{3}{2}+\dfrac{5}{2},\dfrac{3}{2}-\dfrac{5}{2} \\
 & \Rightarrow x=4,-1 \\
\end{align}\]
We graphically solve the problem by plotting the curve in $xy-$plane. We know that $y=a{{x}^{2}}+bx+c$ is the curve of an upward parabola with vertex at $x=\dfrac{-b}{2a}$. Show the points at where the curve $y=a{{x}^{2}}+bx+c$ will cut the $x-$axis will be our solution. So the vertex of the parabola is at $x=\dfrac{-b}{2a}=\dfrac{-3}{2}$ and the plot curve below. \[\]

So we put the curve $y={{x}^{2}}+3x-4$ cuts $x-$axis at the points $\left( -4,0 \right),\left( 1,0 \right)$ . So the solutions of the equation are $x=-4,1$.\[\]

Note: We can alternatively solve by splitting the middle term method if the discriminant $D={{b}^{2}}-4ac$ is a perfect square. Here in this problem the discriminant is $D={{b}^{2}}-4ac={{\left( -3 \right)}^{2}}-4\times 1\times \left( -4 \right)=25$ is perfect square. We can now split the middle tem $-3x$ to have ${{x}^{2}}-4x+x-4=0$ which we factorize to get $\left( x+4 \right)\left( x-1 \right)=0$ from where we get the solution as $x=4,-1$