Can you solve this equation

I.

$$2x^2+4x-5=0$$

$$\small{\text{
$
x_{1,2}=\dfrac{-4 \pm \sqrt{16-4\cdot 2\cdot (-5) } } {2\cdot2 }
= \dfrac{-4 \pm \sqrt{16+40 } } {4}
= \dfrac{-4 \pm \sqrt{56} } {4}
= \dfrac{-4 \pm \sqrt{16\cdot 3.5} } {4}
=-1\pm\sqrt{3.5}
$}}\\
\small{\text{
$
x_1 =-1+\sqrt{3.5} \qquad x_2 =-1-\sqrt{3.5}
$}}$$

II.

$$-4x^2-8x+5=3\\
-4x^2-8x+2=0$$

$$\small{\text{
$
x_{1,2}=\dfrac{ 8 \pm \sqrt{64-4\cdot (-4)\cdot 2 } } {2\cdot (-4)}
= \dfrac{ 8 \pm \sqrt{64+32 } } {-8}
= \dfrac{ 8 \pm \sqrt{96} } {-8}
= \dfrac{ 8 \pm \sqrt{64\cdot 1.5} } {-8}
=-1\mp\sqrt{1.5}
$}}\\
\small{\text{
$
x_1 =-1+\sqrt{1.5} \qquad x_2 =-1-\sqrt{1.5}
$}}$$

III.

$$4x^2-4x+1=0$$

$$\small{\text{
$
x_{1,2}=\dfrac{ 4 \pm \sqrt{16-4\cdot 4 \cdot 1 } } {2\cdot 4 }
= \dfrac{ 4 \pm \sqrt{16-16} } {8}
= \dfrac{ 4 \pm 0 } {8}
= \dfrac{ 4 } {8}
= \dfrac{ 1 } {2}
= 0.5
$}}\\
\small{\text{
$
x_1 = x_2 = 0.5
$}}$$

IV.

$$3x^2+2x+4=0$$

$$\small{\text{
$
x_{1,2}=\dfrac{ -2 \pm \sqrt{4-4\cdot 3 \cdot 4 } } {2\cdot 3 }
= \dfrac{ -2 \pm \sqrt{4-48} } {6}
= \dfrac{ -2 \pm \sqrt{-44} } {6}
$}}\\
\small{\text{
$
\sqrt{-44} $ not real, no solution$
$}}$$

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