If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5x^2+4x-4=0
a = 5; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·5·(-4)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{6}}{2*5}=\frac{-4-4\sqrt{6}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{6}}{2*5}=\frac{-4+4\sqrt{6}}{10} $
| 6(v+4)=36 | | 2^2(x+3)+9−5=32 | | 47=9p-25 | | –6x+3=9 | | 18=r/4+15 | | 2^9x-1=4 | | 4m+7/2=18 | | -1/5p+9/5p=24 | | 5(z+3)=33 | | 12+5n-3=49 | | f-6=(-2f+3(f-2) | | x+40.9=47.57 | | 263=82-v | | 3(x-4)-2(x-5)=8 | | 5x+15=9x/8+20 | | x+4/3=-1 | | (2^(x-8))^(x-10)=8 | | 9x+5-4x-3+x=146. | | y=25,000(0.85)^10 | | y=25,000(0.85)^7 | | 11-(-3+x)=10 | | 2/3x+1=36 | | y=25,000(0.85)^5 | | y=25,000(0.85)^3 | | (x-5)/2)+(2/(x+5))=17/4 | | y=25,000(0.85)^1 | | 23=1/2((3x-1)+(8x+3) | | 8w+5=48 | | 8w+4=48 | | 3m-8m+5=-104 | | 4x+5=x41= | | -6(r+3)=-60 |